General Chemistry 10th Edition by Darrell Ebbing – Test Bank

 

 

To Purchase this Complete Test Bank with Answers Click the link Below

 

https://tbzuiqe.com/product/general-chemistry-10th-edition-by-darrell-ebbing-test-bank/

 

If face any problem or Further information contact us At tbzuiqe@gmail.com

 

Sample Test

Chapter 3 – Calculations with Chemical Formulas and Equations

 

1.    The molecular formula of a particular solid is C8H6O8. Its molecular mass is

2.    A) 352 amu.

3.    B) 118 amu.

4.    C) 364 amu.

5.    D) 230 amu.

6.    E) 226 amu.

 

 

ANS:      D             PTS:       1              DIF:        easy       REF:       3.1

OBJ:       Calculate the formula mass from a formula. (Example 3.1)

TOP:      stoichiometry | mass and moles of substance                    KEY:       molecular mass

MSC:     general chemistry

 

2.    The hydrocarbon heptane has the structural formula CH3(CH2)5CH3.  What is the molecular mass of this hydrocarbon?

3.    A) 2 amu.

4.    B) 009980 amu.

5.    C) 17 amu.

6.    D) 01160 amu.

7.    E) 2 amu.

 

 

ANS:      A             PTS:       1              DIF:        easy       REF:       3.1

OBJ:       Calculate the formula mass from a formula.

TOP:      stoichiometry | mass and moles of substance

 

3.    A single molecule of polystryene has the repeating unit -[CH2CH(C6H5)]n-, where n is the number of repeating units.  What is the value of n if the molecular mass of a single polymer chain is   amu?

4.    A)

5.    B)

6.    C)

7.    D)

8.    E)

 

 

ANS:      A             PTS:       1              DIF:        moderate            REF:       3.1

OBJ:       Calculate the formula mass from a formula.

TOP:      stoichiometry | mass and moles of substance

 

4.    The formula mass of aluminum oxalate, Al2(C2O4)3, is

5.    A) 140 amu.

6.    B) 318 amu.

7.    C) 283 amu.

8.    D) 213 amu.

9.    E) 185 amu.

 

 

ANS:      B             PTS:       1              DIF:        easy       REF:       3.1

OBJ:       Calculate the formula mass from a formula. (Example 3.1)

TOP:      stoichiometry | mass and moles of substance                    KEY:       formula mass

MSC:     general chemistry

 

5.    The formula mass of zinc acetate trihydrate, Zn(CH3COO)2 • 3H2O, is

6.    A) 321 amu.

7.    B) 184 amu.

8.    C) 268 amu.

9.    D) 238 amu.

10.  E) 114 amu.

 

 

ANS:      D             PTS:       1              DIF:        easy       REF:       3.1

OBJ:       Calculate the formula mass from a formula. (Example 3.1)

TOP:      stoichiometry | mass and moles of substance                    KEY:       formula mass

MSC:     general chemistry

 

6.    The fully hydrated form of sodium sulfate is the decahydrate, Na2SO4 • 10H2O.  When heated the hydrated salt loses water.  How many water molecules are found per formula unit in a partially dehydrated sample of sodium sulfate with a formula mass of 160.1 amu (i.e. find n for  Na2SO4 • nH2O)?

7.    A) 1 waters.

8.    B) 5 waters.

9.    C) 3 waters.

10.  D) 4 waters.

11.  E) 7 waters.

 

 

ANS:      A             PTS:       1              DIF:        moderate            REF:       3.1

OBJ:       Calculate the formula mass from a formula.

TOP:      stoichiometry | mass and moles of substance

 

7.    What is the molecular mass of cycloheptane, C7H14?

8.    A) 02 amu

9.    B) 19 amu

10.  C) 08 amu

11.  D) 19 amu

12.  E) 12 amu

 

 

ANS:      D             PTS:       1              DIF:        easy       REF:       3.1

OBJ:       Calculate the formula mass from a formula. (Example 3.1)

TOP:      stoichiometry | mass and moles of substance                    KEY:       molecular mass

MSC:     general chemistry

 

8.    What is the formula mass of strontium phosphate, Sr3(PO4)2?

9.    A) 83 amu

10.  B) 83 amu

11.  C) 59 amu

12.  D) 66 amu

13.  E) 80 amu

 

 

ANS:      E              PTS:       1              DIF:        easy       REF:       3.1

OBJ:       Calculate the formula mass from a formula. (Example 3.1)

TOP:      stoichiometry | mass and moles of substance                    KEY:       formula mass

MSC:     general chemistry

 

9.    What is the molecular mass of the hydrocarbon styrene (shown in the figure)?

 

 

104.          A) 1 amu.

105.          B) 1 amu.

106.          C) 1 amu.

107.          D) 0 amu.

108.          E) 1 amu.

 

 

ANS:      A             PTS:       1              DIF:        easy       REF:       3.1

OBJ:       Calculate the formula mass from molecular models. (Example 3.2)

TOP:      stoichiometry | mass and moles of substance                    KEY:       molecular mass

MSC:     general chemistry

 

10.  What is the molar mass of ammonium sulfite, (NH4)2SO3?

11.  A) 98 g/mol

12.  B) 116 g/mol

13.  C) 55 g/mol

14.  D) 180 g/mol

15.  E) 84 g/mol

 

 

ANS:      B             PTS:       1              DIF:        easy       REF:       3.2

OBJ:       Understand how the molar mass is related to the formula weight of a substance.

TOP:      stoichiometry | mass and moles of substance                    KEY:       formula mass

MSC:     general chemistry

 

11.  What is the molar mass of zinc sulfate heptahydrate, ZnSO4 • 7H2O?

12.  A) g/mol

13.  B) 288 g/mol

14.  C) 384 g/mol

15.  D) 162 g/mol

16.  E) 582 g/mol

 

 

ANS:      B             PTS:       1              DIF:        easy       REF:       3.2

OBJ:       Understand how the molar mass is related to the formula weight of a substance.

TOP:      stoichiometry | mass and moles of substance                    KEY:       formula mass

MSC:     general chemistry

 

12.  Plastic wrap can be made from poly(vinylidene chloride).  A single poly(vinylidene chloride) strand has the general formula -(CH2CHCl)n-,  where n ranges from 10,000 to 100,000.  What is the molar mass of a single poly(vinylidene chloride) molecule containing   repeating units?

13.  A) g/mol

14.  B) g/mol

15.  C) g/mol

16.  D) g/mol

17.  E) g/mol

 

 

ANS:      A             PTS:       1              DIF:        easy       REF:       3.2

OBJ:       Understand how the molar mass is related to the formula weight of a substance.

TOP:      stoichiometry | mass and moles of substance

 

13.  The dicarboxylic acid potassium hydrogen pthalate (shown in the figure) is used to standardize solutions of strong base.  What is the molar mass of this compound?

 

 

204.          A) 2 g/mol

205.          B) 2 g/mol

206.          C) 9 g/mol

207.          D) 08 g/mol

208.          E) 2 g/mol

 

 

ANS:      A             PTS:       1              DIF:        easy       REF:       3.2

OBJ:       Understand how the molar mass is related to the formula weight of a substance.

TOP:      stoichiometry | mass and moles of substance

 

14.  What is the molar mass of the solid C8H12N2O4?

15.  A) 172 g/mol

16.  B) 136 g/mol

17.  C) 200 g/mol

18.  D) 106 g/mol

19.  E) 188 g/mol

 

 

ANS:      C             PTS:       1              DIF:        easy       REF:       3.2

OBJ:       Understand how the molar mass is related to the formula weight of a substance.

TOP:      stoichiometry | mass and moles of substance                    KEY:       molecular mass

MSC:     general chemistry

 

15.  A 1.067 g sample of an element contains 5.062 ´ 1021 atoms. What is the element symbol?

16.  A) I

17.  B) Ag

18.  C) La

19.  D) Pd

20.  E) Rh

 

 

ANS:      A             PTS:       1              DIF:        moderate            REF:       3.2

OBJ:       Understand how the molar mass is related to the formula weight of a substance.

TOP:      stoichiometry | determining chemical formulas

 

16.  Monodisperse polyacrylonitrile contains molecules with the general formula -(CH2CHCN)n-, where n is typically greater than 10,000.  Given that a sample of monodisperse polyacrilonitrile weighs 197.4 g and contains   molecules of -(CH2CHCN)n-, calculate n.

17.  A)

18.  B)

19.  C)

20.  D)

21.  E)

 

 

ANS:      A             PTS:       1              DIF:        difficult REF:       3.2

OBJ:       Understand how the molar mass is related to the formula weight of a substance.

TOP:      stoichiometry | mass and moles of substance

 

17.  Monodisperse polyacrylonitrile contains molecules with the general formula -(CH2CHCN)n-, where n is typically greater than 10,000.  Given that a sample of monodisperse polyacrylonitrile weighs 252.6 g and contains   molecules of -(CH2CHCN)n-, what is the molar mass of the polymer?

18.  A) g/mol

19.  B) g/mol

20.  C) g/mol

21.  D) g/mol

22.  E) g/mol

 

 

ANS:      E              PTS:       1              DIF:        difficult REF:       3.2

OBJ:       Understand how the molar mass is related to the formula weight of a substance.

TOP:      stoichiometry | mass and moles of substance

 

18.  An atom of an element weighs 3.00 ´ 10–23 g. What is the atomic mass of this element in atomic mass units?

19.  A) 4 amu

20.  B) 5 amu

21.  C) 9 amu

22.  D) 1 amu

23.  E) 3 amu

 

 

ANS:      D             PTS:       1              DIF:        easy       REF:       3.2

OBJ:       Calculate the mass of atoms and molecules. (Example 3.3)

TOP:      stoichiometry | mass and moles of substance

KEY:       mole | mole calculations               MSC:     general chemistry

 

19.  What is the mass in grams of one propene, C3H6, molecule?

20.  A) 99 ´ 10–23 g

21.  B) 53 ´ 1025 g

22.  C) 0 g

23.  D) 0 g

24.  E) 99 ´ 10–23 g

 

 

ANS:      A             PTS:       1              DIF:        easy       REF:       3.2

OBJ:       Calculate the mass of atoms and molecules. (Example 3.3)

TOP:      stoichiometry | mass and moles of substance

KEY:       mole | mole calculations               MSC:     general chemistry

 

20.  What is the mass of oxygen in grams found in one molecule of the compound C7H8O4?

21.  A) g

22.  B) g

23.  C) g

24.  D) g

25.  E) 156 g

 

 

ANS:      A             PTS:       1              DIF:        moderate            REF:       3.2

OBJ:       Calculate the mass of atoms and molecules. (Example 3.3)

TOP:      stoichiometry | mass and moles of substance

 

21.  Which of the following compounds contains the largest number of atoms?

22.  A) 0 mol of K2S

23.  B) 0 mol of NH3

24.  C) 0 mol of H2SO4

25.  D) 0 mol of HCl

26.  E) 0 mol of CH3COCl

 

 

ANS:      C             PTS:       1              DIF:        easy       REF:       3.2

OBJ:       Perform calculations using the mole.

TOP:      stoichiometry | mass and moles of substance

KEY:       mole | mole calculations               MSC:     general chemistry

 

22.  How many atoms of carbon are there in 0.51 mol of procaine, C13H20N2O2, a “pain killer” used by dentists?

23.  A) 6 ´ 1023

24.  B) 3 ´ 1024

25.  C) 0 ´ 1024

26.  D) 1 ´ 1023

27.  E) 6 ´ 1024

 

 

ANS:      C             PTS:       1              DIF:        easy       REF:       3.2

OBJ:       Perform calculations using the mole.

TOP:      stoichiometry | mass and moles of substance

KEY:       mole | mole calculations               MSC:     general chemistry

 

23.  A sample of ammonium phosphite, (NH4)3PO3, contains 0.909 mol of hydrogen atoms. The number of moles of oxygen atoms in the sample is

24.  A) 227 mol.

25.  B) 0909 mol.

26.  C) 300 mol.

27.  D) 227 mol.

28.  E) 00 mol.

 

 

ANS:      A             PTS:       1              DIF:        easy       REF:       3.2

OBJ:       Perform calculations using the mole.

TOP:      stoichiometry | mass and moles of substance

 

24.  A sample of gallium(III) sulfite, Ga2(SO3)3, contains 1.95 mol of sulfite ions. The number of moles of gallium(III) ions in the sample is

25.  A) 30 mol.

26.  B) 92 mol.

27.  C) 95 mol.

28.  D) 84 mol.

29.  E) 90 mol.

 

 

ANS:      A             PTS:       1              DIF:        easy       REF:       3.2

OBJ:       Perform calculations using the mole.

TOP:      stoichiometry | mass and moles of substance

 

25.  Sorbose, C6H12O6, is used in making vitamin C.  A sorbose sample containing 96.0 g of carbon atoms also contains ____ g of hydrogen atoms.

26.  A) 2

27.  B) 1

28.  C) 15 ´ 103

29.  D) 0

30.  E) 99

 

 

ANS:      B             PTS:       1              DIF:        moderate            REF:       3.2

OBJ:       Perform calculations using the mole.

TOP:      stoichiometry | mass and moles of substance

KEY:       mole | mole calculations               MSC:     general chemistry

 

26.  Consider the following three samples.

27.  A sample containing 180 g glucose (C6H12O6)

28.  A sample containing 90 g glucose and 90 g fructose (C6H12O6)

29.  A sample containing 180 g fructose

Which statement is correct?

1.    A) All three samples have the same number of hydrogen atoms.

2.    B) Both samples A and C have the same number of hydrogen atoms, but more than in sample B.

3.    C) Sample B has more hydrogen atoms than sample A or sample C.

4.    D) Sample C has more hydrogen atoms than sample A or sample B

5.    E) Sample A has more hydrogen atoms than sample B or sample C.

 

 

ANS:      A             PTS:       1              DIF:        difficult REF:       3.2

OBJ:       Perform calculations using the mole.

TOP:      stoichiometry | determining chemical formulas

KEY:       mole | mole calculations               MSC:     general chemistry

 

27.  Styrene’s empirical formula is CH. What mass of styrene contains 7.89 ´ 1021 atoms of hydrogen?  The molar mass of styrene is 104 g/mol.

28.  A) 105 g

29.  B) 170 g

30.  C) 36 g

31.  D) 157 g

32.  E) 0131 g

 

 

ANS:      B             PTS:       1              DIF:        moderate            REF:       3.2

OBJ:       Convert from moles of substance to grams of substance. (Example 3.4)

TOP:      stoichiometry | mass and moles of substance

KEY:       mole | mole calculations               MSC:     general chemistry

 

28.  What is the mass of oxygen atoms in 0.305 mol Fe(CO)5?

29.  A) 0 g

30.  B) 7 g

31.  C) 4 g

32.  D) 88 g

33.  E) 3 g

 

 

ANS:      C             PTS:       1              DIF:        moderate            REF:       3.2

OBJ:       Convert from moles of substance to grams of substance. (Example 3.4)

TOP:      stoichiometry | mass and moles of substance

KEY:       mole | mole calculations               MSC:     general chemistry

 

29.  What is the mass in grams of 0.699 mol of glucose, C6H12O6?

30.  A) 00388 g

31.  B) 1 g

32.  C) 126 g

33.  D) 0 g

34.  E) 258 g

 

 

ANS:      C             PTS:       1              DIF:        easy       REF:       3.2

OBJ:       Convert from moles of substance to grams of substance. (Example 3.4)

TOP:      stoichiometry | mass and moles of substance

KEY:       mole | mole calculations               MSC:     general chemistry

 

30.  Which one of the following samples has the greatest mass?

31.  A) 39 mol of camphor, C10H16O

32.  B) 1 mol of ammonia, NH3

33.  C) 2 mol of krypton, Kr

34.  D) 6 mol of iodine vapor, I2

35.  E) 8 mol of formaldehyde, CH2O

 

 

ANS:      D             PTS:       1              DIF:        moderate            REF:       3.2

OBJ:       Convert from moles of substance to grams of substance. (Example 3.4)

TOP:      stoichiometry | mass and moles of substance

KEY:       mole | mole calculations               MSC:     general chemistry

 

31.  Which of the following contains the greatest mass of oxygen atoms?

32.  A) 6 mol CoSO4 • 7H2O

33.  B) 3 mol KHSO4

34.  C) 2 mol K2Cr2O7

35.  D) 3 mol H2O2

36.  E) 3 mol Na2S2O3

 

 

ANS:      B             PTS:       1              DIF:        easy       REF:       3.2

OBJ:       Convert from moles of substance to grams of substance. (Example 3.4)

TOP:      stoichiometry | mass and moles of substance

KEY:       mole | mole calculations               MSC:     general chemistry

 

32.  Calculate the number of moles of bromine present in 14.5 mL of Br2(l), whose density is 3.12 g/mL.

33.  A) 53 mol

34.  B) 181 mol

35.  C) 566 mol

36.  D) 091 mol

37.  E) 283 mol

 

 

ANS:      E              PTS:       1              DIF:        moderate            REF:       3.2

OBJ:       Convert from grams of substance to moles of substance. (Example 3.5)

TOP:      stoichiometry | mass and moles of substance

KEY:       mole | mole calculations               MSC:     general chemistry

 

33.  How many moles of hexachlorobenzene, C6Cl6, are in 4.45 g of C6Cl6?

34.  A) 0208 mol

35.  B) 27 ´ 103 mol

36.  C) 0618 mol

37.  D) 0156 mol

38.  E) 0322 mol

 

 

ANS:      D             PTS:       1              DIF:        easy       REF:       3.2

OBJ:       Convert from grams of substance to moles of substance. (Example 3.5)

TOP:      stoichiometry | mass and moles of substance

KEY:       mole | mole calculations               MSC:     general chemistry

 

34.  How many moles of iron atoms are contained in 4.39 g of iron?

35.  A) 245 mol

36.  B) 0579 mol

37.  C) 0786 mol

38.  D) 122 mol

39.  E) 169 mol

 

 

ANS:      C             PTS:       1              DIF:        easy       REF:       3.2

OBJ:       Convert from grams of substance to moles of substance. (Example 3.5)

TOP:      stoichiometry | mass and moles of substance

KEY:       mole | mole calculations               MSC:     general chemistry

 

35.  How many moles of pentane, C5H12, are contained in a 11-g sample?

36.  A) 18 mol

37.  B) 15 mol

38.  C) 26 mol

39.  D) 4 mol

40.  E) 1 mol

 

 

ANS:      B             PTS:       1              DIF:        easy       REF:       3.2

OBJ:       Convert from grams of substance to moles of substance. (Example 3.5)

TOP:      stoichiometry | mass and moles of substance

KEY:       mole | mole calculations               MSC:     general chemistry

 

36.  Sodium cyclamate, C6H11NHSO3Na, was used at one time as an artificial sweetener. C6H11NHSO3Na has a molecular mass of 201.2 g/mol. How many moles of sodium cyclamate are contained in a 67.6-g sample?

37.  A) 211 mol

38.  B) 193 mol

39.  C) 336 mol

40.  D) 307 mol

41.  E) 13,600 mol

 

 

ANS:      C             PTS:       1              DIF:        easy       REF:       3.2

OBJ:       Convert from grams of substance to moles of substance. (Example 3.5)

TOP:      stoichiometry | mass and moles of substance

KEY:       mole | mole calculations               MSC:     general chemistry

 

37.  How many moles of silver are contained in 7.00 kg of silver?

38.  A) 9 mol

39.  B) 9 ´ 101 mol

40.  C) 9 ´ 10–3 mol

41.  D) 9 ´ 103 mol

42.  E) 9 ´ 10–1 mol

 

 

ANS:      A             PTS:       1              DIF:        easy       REF:       3.2

OBJ:       Convert from grams of substance to moles of substance. (Example 3.5)

TOP:      stoichiometry | mass and moles of substance

KEY:       mole | mole calculations               MSC:     general chemistry

 

38.  A 0.0103-mol sample of urea, NH2CONH2, contains

39.  A) 02 ´ 1023 molecules.

40.  B) 48 ´ 1022 molecules.

41.  C) 96 ´ 1022 atoms.

42.  D) 48 ´ 1023 atoms.

43.  E) 03 ´ 1024 atoms.

 

 

ANS:      C             PTS:       1              DIF:        easy       REF:       3.2

OBJ:       Calculate the number of molecules in a given mass of substance. (Example 3.6)

TOP:      stoichiometry | mass and moles of substance

KEY:       mole | mole calculations               MSC:     general chemistry

 

39.  How many molecules are there in 90.0 g of butylene glycol, HO(CH2)4OH?

40.  A) 1

41.  B) (6.02 ´ 1023)/90.0

42.  C) 0

43.  D) 0 ´ (6.02 ´ 1023)

44.  E) 02 ´ 1023

 

 

ANS:      E              PTS:       1              DIF:        easy       REF:       3.2

OBJ:       Calculate the number of molecules in a given mass of substance. (Example 3.6)

TOP:      stoichiometry | mass and moles of substance

KEY:       mole | mole calculations               MSC:     general chemistry

 

40.  How many molecules are there in 192 g of citric acid, C6H8O7?

41.  A) (6.02 ´ 1023) / 192

42.  B) 192

43.  C) 02 ´ 1023

44.  D) 0

45.  E) 192 ´ (6.02 ´ 1023)

 

 

ANS:      C             PTS:       1              DIF:        easy       REF:       3.2

OBJ:       Calculate the number of molecules in a given mass of substance. (Example 3.6)

TOP:      stoichiometry | mass and moles of substance

KEY:       mole | mole calculations               MSC:     general chemistry

 

41.  In 0.358 mol of trimellitic acid, C6H3(COOH)3, there are

42.  A) 16 ´ 1022 molecules.

43.  B) 62 ´ 1024 molecules.

44.  C) 47 ´ 1023 oxygen atoms.

45.  D) 94 ´ 1024 carbon atoms.

46.  E) 59 ´ 1023 hydrogen atoms.

 

 

ANS:      D             PTS:       1              DIF:        moderate            REF:       3.2

OBJ:       Calculate the number of molecules in a given mass of substance. (Example 3.6)

TOP:      stoichiometry | mass and moles of substance

KEY:       mole | mole calculations               MSC:     general chemistry

 

42.  How many molecules are there in 2.80 kg of hydrazine, N2H4?

43.  A) 75 ´ 1023

44.  B) 27 ´ 1025

45.  C) 88 ´ 1022

46.  D) 80 ´ 1026

47.  E) 65 ´ 1021

 

 

ANS:      B             PTS:       1              DIF:        moderate            REF:       3.2

OBJ:       Calculate the number of molecules in a given mass of substance. (Example 3.6)

TOP:      stoichiometry | mass and moles of substance

KEY:       mole | mole calculations               MSC:     general chemistry

 

43.  How many molecules are there in 2.43 mg of mannose, C6H12O6, which is a sweet-tasting sugar that has a bitter aftertaste?

44.  A) 46 ´ 1021

45.  B) 48 ´ 1021

46.  C) 31 ´ 1018

47.  D) 04 ´ 1024

48.  E) 13 ´ 1018

 

 

ANS:      E              PTS:       1              DIF:        moderate            REF:       3.2

OBJ:       Calculate the number of molecules in a given mass of substance. (Example 3.6)

TOP:      stoichiometry | mass and moles of substance

KEY:       mole | mole calculations               MSC:     general chemistry

 

44.  How many atoms are present in 463 g of KPF6 (MM = 184.1 g/mol)?

45.  A) 23 ´ 1025

46.  B) 51 ´ 1021

47.  C) 51 ´ 1021

48.  D) 13 ´ 1026

49.  E) 21 ´ 1025

 

 

ANS:      E              PTS:       1              DIF:        moderate            REF:       3.2

OBJ:       Calculate the number of molecules in a given mass of substance. (Example 3.6)

TOP:      stoichiometry | mass and moles of substance

KEY:       mole | mole calculations               MSC:     general chemistry

 

45.  Styrene’s empirical formula is CH. When it is heated to 200°C, it is converted into a polymer, polystyrene, which has excellent insulating properties. What mass of styrene contains 9.75 ´ 1021 molecules of styrene?  The molar mass of styrene is 104 g/mol.

46.  A) 0162 g

47.  B) 68 g

48.  C) 24 g

49.  D) 210 g

50.  E) 130 g

 

 

ANS:      B             PTS:       1              DIF:        difficult REF:       3.2

OBJ:       Calculate the number of molecules in a given mass of substance. (Example 3.6)

TOP:      stoichiometry | mass and moles of substance

KEY:       mole | mole calculations               MSC:     general chemistry

 

46.  In 0.500 mol of dimethylhydrazine, (CH3)2N2H2, there are

47.  A) 01 ´ 1024 molecules.

48.  B) 51 ´ 1023 atoms.

49.  C) 01 ´ 1022 molecules.

50.  D) 61 ´ 1024 atoms.

51.  E) 81 ´ 1024 atoms.

 

 

ANS:      D             PTS:       1              DIF:        moderate            REF:       3.2

OBJ:       Calculate the number of molecules in a given mass of substance. (Example 3.6)

TOP:      stoichiometry | mass and moles of substance

KEY:       mole | mole calculations               MSC:     general chemistry

 

47.  Which is a reasonable mass corresponding to 1020  molecules of a substance?

48.  A) 100 g

49.  B) 100 µg

50.  C) 100 ng

51.  D) 100 mg

52.  E) 100 kg

 

 

ANS:      D             PTS:       1              DIF:        moderate            REF:       3.2

OBJ:       Calculate the number of molecules in a given mass of substance. (Example 3.6)

TOP:      stoichiometry | mass and moles of substance

KEY:       mole | mole calculations               MSC:     general chemistry

 

48.  How many aluminum atoms are there in 52 g of Al2S3?

49.  A) 2 ´ 1023

50.  B) 6 ´ 1021

51.  C) 1 ´ 1023

52.  D) 1 ´ 1021

53.  E) 3 ´ 1023

 

 

ANS:      A             PTS:       1              DIF:        moderate            REF:       3.2

OBJ:       Calculate the number of molecules in a given mass of substance. (Example 3.6)

TOP:      stoichiometry | mass and moles of substance

KEY:       mole | mole calculations               MSC:     general chemistry

 

49.  Which one of the following contains 2.41 ´ 1024 atoms?

50.  A) 0 g C2H2

51.  B) 0 g O2

52.  C) 0 g CH4

53.  D) 168 g N2

54.  E) 0 g He

 

 

ANS:      E              PTS:       1              DIF:        moderate            REF:       3.2

OBJ:       Calculate the number of molecules in a given mass of substance. (Example 3.6)

TOP:      stoichiometry | mass and moles of substance

KEY:       mole | mole calculations               MSC:     general chemistry

 

50.  A sample of 496 g of white phosphorus, P4, contains the same number of atoms as

51.  A) 192 g of ozone (O3).

52.  B) 0 g of nitrogen (N2).

53.  C) 0 g of sodium.

54.  D) 120 g of formaldehyde (CH2O).

55.  E) 128 g of oxygen (O2).

 

 

ANS:      D             PTS:       1              DIF:        moderate            REF:       3.2

OBJ:       Calculate the number of molecules in a given mass of substance. (Example 3.6)

TOP:      stoichiometry | mass and moles of substance

KEY:       mole | mole calculations               MSC:     general chemistry

 

51.  The total number of oxygen atoms in 1.76 g of CaCO3 (MM = 100.0 g/mol) is

52.  A) 05 ´ 1023.

53.  B) 24 ´ 1022.

54.  C) 18 ´ 1022.

55.  D) 75 ´ 1023.

56.  E) 30 ´ 1022.

 

 

ANS:      C             PTS:       1              DIF:        moderate            REF:       3.2

OBJ:       Calculate the number of molecules in a given mass of substance. (Example 3.6)

TOP:      stoichiometry | mass and moles of substance

KEY:       mole | mole calculations               MSC:     general chemistry

 

52.  A sample of 336 g of ozone, O3, contains the same number of atoms as

53.  A) 336 g of oxygen (O2).

54.  B) 2 g of hydrogen (H2).

55.  C) 266 g of fluorine (F2).

56.  D) 189 g of aluminum (Al).

57.  E) 411 g of nickel (Ni).

 

 

ANS:      A             PTS:       1              DIF:        moderate            REF:       3.2

OBJ:       Calculate the number of molecules in a given mass of substance. (Example 3.6)

TOP:      stoichiometry | mass and moles of substance

KEY:       mole | mole calculations               MSC:     general chemistry

 

53.  Which of the following samples contains the smallest number of molecules?

54.  A) 00 g of TNT, C7H5N3O6

55.  B) 00 g of benzene, C6H6

56.  C) 00 g of glucose, C6H12O6

57.  D) 00 g of naphthalene, C10H8

58.  E) 00 g of formaldehyde, CH2O

 

 

ANS:      A             PTS:       1              DIF:        moderate            REF:       3.2

OBJ:       Calculate the number of molecules in a given mass of substance. (Example 3.6)

TOP:      stoichiometry | mass and moles of substance

KEY:       mole | mole calculations               MSC:     general chemistry

 

54.  Which of the following samples contains the largest number of atoms?

55.  A) 1 g N2

56.  B) 1 g Li

57.  C) 1 g Cl2

58.  D) 1 g P4

59.  E) 1 g Mg

 

 

ANS:      B             PTS:       1              DIF:        moderate            REF:       3.2

OBJ:       Calculate the number of molecules in a given mass of substance. (Example 3.6)

TOP:      stoichiometry | mass and moles of substance

KEY:       mole | mole calculations               MSC:     general chemistry

 

55.  Which of the following samples contains the largest number of molecules?

56.  A) g Pb

57.  B) g Cl2

58.  C) g Kr

59.  D) g O2

60.  E) g S8

 

 

ANS:      D             PTS:       1              DIF:        moderate            REF:       3.2

OBJ:       Calculate the number of molecules in a given mass of substance. (Example 3.6)

TOP:      stoichiometry | mass and moles of substance

KEY:       mole | mole calculations               MSC:     general chemistry

 

56.  In 1928, 1.0 g of rhenium, Re, was isolated from 660 kg of the ore molybenite. The percent by mass of this element in the molybenite was

57.  A) 66 %.

58.  B) 15 %.

59.  C) 5 ´ 10–4 %.

60.  D) 6 ´ 10–3 %.

61.  E) 5 ´ 10–4 %.

 

 

ANS:      E              PTS:       1              DIF:        easy       REF:       3.3

OBJ:       Calculate the percentage composition of the elements in a compound. (Example 3.7)

TOP:      stoichiometry | determining chemical formulas KEY:       mass percentage

MSC:     general chemistry

 

57.  An ore sample is found to contain 24.1 g of mercury and 50.7 g waste rock (gangue).  What is the percent by mass of mercury in the ore?

58.  A) 2 %

59.  B) 4 %

60.  C) 322 %

61.  D) 474 %

62.  E) 74 %

 

 

ANS:      A             PTS:       1              DIF:        easy       REF:       3.3

OBJ:       Calculate the percentage composition of the elements in a compound. (Example 3.7)

TOP:      stoichiometry | determining chemical formulas

 

58.  An ore sample with a mass of 68.0 g is found to contain 15.5% by mass nickel.  What mass of nickel is contained in the ore?

59.  A) 5 g

60.  B) 7 g

61.  C) 55 g

62.  D) 5 g

63.  E) 546 g

 

 

ANS:      A             PTS:       1              DIF:        easy       REF:       3.3

OBJ:       Calculate the percentage composition of the elements in a compound. (Example 3.7)

TOP:      stoichiometry | determining chemical formulas

 

59.  What is the percent by mass oxygen in (NH4)2SO3?

60.  A) 3 %

61.  B) 7 %

62.  C) 0 %

63.  D) 0 %

64.  E) 00 %

 

 

ANS:      A             PTS:       1              DIF:        easy       REF:       3.3

OBJ:       Calculate the percentage composition of the elements in a compound. (Example 3.7)

TOP:      stoichiometry | determining chemical formulas

 

60.  What is the percentage by mass of hydrogen in the insecticide Lindane, C6H6Cl6?

61.  A) 0 %

62.  B) 20 %

63.  C) 2 %

64.  D) 80 %

65.  E) 08 %

 

 

ANS:      E              PTS:       1              DIF:        easy       REF:       3.3

OBJ:       Calculate the percentage composition of the elements in a compound. (Example 3.7)

TOP:      stoichiometry | determining chemical formulas

 

61.  The mineral leadhillite, which is essentially Pb4(SO4)(CO3)2(OH)2 (FW = 1079 g/mol), contains ____% hydrogen by mass.

62.  A) 81

63.  B) 1868

64.  C) 226

65.  D) 79

66.  E) 972

 

 

ANS:      B             PTS:       1              DIF:        easy       REF:       3.3

OBJ:       Calculate the percentage composition of the elements in a compound. (Example 3.7)

TOP:      stoichiometry | determining chemical formulas KEY:       mass percentage

MSC:     general chemistry

 

62.  Which of the following compounds has the highest percentage of hydrogen atoms by mass?

63.  A) CH3COOH

64.  B) C2H5OH

65.  C) CH3OH

66.  D) H2CO3

67.  E) H2C2O4

 

 

ANS:      B             PTS:       1              DIF:        easy       REF:       3.3

OBJ:       Calculate the percentage composition of the elements in a compound. (Example 3.7)

TOP:      stoichiometry | determining chemical formulas KEY:       mass percentage

MSC:     general chemistry

 

63.  Which of the following compounds has the highest percentage of nitrogen by mass?

64.  A) (NH4)2SO3

65.  B) NaNO3

66.  C) N2Cl4

67.  D) NH4NO2

68.  E) HNO3

 

 

ANS:      D             PTS:       1              DIF:        easy       REF:       3.3

OBJ:       Calculate the percentage composition of the elements in a compound. (Example 3.7)

TOP:      stoichiometry | determining chemical formulas KEY:       mass percentage

MSC:     general chemistry

 

64.  Which of the following compounds has the same percentage of carbon and hydrogen by mass as cyclohexane, C6H12?

65.  A) C6H14, hexane

66.  B) C5H10, pentene

67.  C) C6H10, cyclohexene

68.  D) C6H6, benzene

69.  E) C6H12O6, glucose

 

 

ANS:      B             PTS:       1              DIF:        easy       REF:       3.3

OBJ:       Calculate the percentage composition of the elements in a compound. (Example 3.7)

TOP:      stoichiometry | determining chemical formulas KEY:       mass percentage

MSC:     general chemistry

 

65.  What is the mass percentage of carbon in the compound C6H6O2?

66.  A) 5 %

67.  B) 9 %

68.  C) 5 %

69.  D) 1 %

70.  E) 3 %

 

 

ANS:      C             PTS:       1              DIF:        easy       REF:       3.3

OBJ:       Calculate the percentage composition of the elements in a compound. (Example 3.7)

TOP:      stoichiometry | determining chemical formulas KEY:       mass percentage

MSC:     general chemistry

 

66.  What is the percentage by mass of hydrogen in ammonium phosphate, (NH4)3PO4?

67.  A) 11 %

68.  B) 06 %

69.  C) 0 %

70.  D) 0 %

71.  E) 00 %

 

 

ANS:      A             PTS:       1              DIF:        easy       REF:       3.3

OBJ:       Calculate the percentage composition of the elements in a compound. (Example 3.7)

TOP:      stoichiometry | determining chemical formulas

 

67.  A crystal of the mineral troegerite, (UO2)3(AsO4)2 • 12H2O (FM = 1304 amu), contains ____% arsenic by mass.

68.  A) 4

69.  B) 8

70.  C) 0

71.  D) 4

72.  E) 5

 

 

ANS:      E              PTS:       1              DIF:        easy       REF:       3.3

OBJ:       Calculate the percentage composition of the elements in a compound. (Example 3.7)

TOP:      stoichiometry | determining chemical formulas KEY:       mass percentage

MSC:     general chemistry

 

68.  How many grams of hydrogen atoms are present in 18.4 g of water?

69.  A) 1 g

70.  B) 02 g

71.  C) 06 g

72.  D) 96 g

73.  E) 3 g

 

 

ANS:      C             PTS:       1              DIF:        easy       REF:       3.3

OBJ:       Calculate the mass of an element in a given mass of compound. (Example 3.8)

TOP:      stoichiometry | mass and moles of substance

KEY:       mole | mole calculations               MSC:     general chemistry

 

69.  How many grams of potassium are present in 21.6 g of K2Cr2O7?

70.  A) 74 g

71.  B) 105 g

72.  C) 87 g

73.  D) 2 g

74.  E) 8 g

 

 

ANS:      A             PTS:       1              DIF:        moderate            REF:       3.3

OBJ:       Calculate the mass of an element in a given mass of compound. (Example 3.8)

TOP:      stoichiometry | determining chemical formulas KEY:       mass percentage

MSC:     general chemistry

 

70.  NaHCO3 is the active ingredient in baking soda. How many grams of oxygen are present in 0.67 g of NaHCO3?

71.  A) 024 g

72.  B) 128 g

73.  C) 98 ´ 103 g

74.  D) 043 g

75.  E) 38 g

 

 

ANS:      E              PTS:       1              DIF:        easy       REF:       3.3

OBJ:       Calculate the mass of an element in a given mass of compound. (Example 3.8)

TOP:      stoichiometry | determining chemical formulas KEY:       mass percentage

MSC:     general chemistry

 

71.  Which of the following contains the greatest mass of bromine atoms?

72.  A) 0 g of KBr

73.  B) 0 g of Br2

74.  C) 076 mol of KBr

75.  D) 092 mol of Br2

76.  E) 0 g of NaBrO3

 

 

ANS:      B             PTS:       1              DIF:        moderate            REF:       3.3

OBJ:       Calculate the mass of an element in a given mass of compound. (Example 3.8)

TOP:      stoichiometry | determining chemical formulas KEY:       mass percentage

MSC:     general chemistry

 

72.  The amount of calcium in a 15.0-g sample was determined by converting the calcium to calcium oxalate, CaC2O4.  The CaC2O4 weighed 12.6 g.  What is the percent of calcium in the original sample?

73.  A) 8 %

74.  B) 3 %

75.  C) 8 %

76.  D) 7 %

77.  E) 0 %

 

 

ANS:      B             PTS:       1              DIF:        difficult REF:       3.3

OBJ:       Calculate the mass of an element in a given mass of compound. (Example 3.8)

TOP:      stoichiometry | mass and moles of substance

KEY:       mole | mole calculations               MSC:     general chemistry

 

73.  A compound containing only carbon, hydrogen, and oxygen is subjected to elemental analysis.  Upon complete combustion, a 0.1804-g sample of the compound produced 0.3051 g of CO2 and 0.1249 g of H2O.  What is the empirical formula of the compound?

74.  A) C3H6O3

75.  B) C3H3O

76.  C) C4H8O3

77.  D) C2H2O

78.  E) CH2O3

 

 

ANS:      C             PTS:       1              DIF:        moderate            REF:       3.4

OBJ:       Calculate the percentage of C, H, and O from combustion data. (Example 3.9)

TOP:      stoichiometry | determining chemical formulas KEY:       elemental analysis

MSC:     general chemistry

 

74.  A 3.075 g sample of a compound containing only carbon, hydrogen, and oxygen is burned in an excess of dioxygen, producing 6.990 g CO2 and 2.862 g H2O.  What mass of oxygen is contained in the original sample?

75.  A) 8472 g

76.  B) 167 g

77.  C) 915 g

78.  D) 129 g

79.  E) 2134 g

 

 

ANS:      A             PTS:       1              DIF:        moderate            REF:       3.4

OBJ:       Calculate the percentage of C, H, and O from combustion data. (Example 3.9)

TOP:      stoichiometry | determining chemical formulas

 

75.  A 4.043 g sample of a compound containing only carbon, hydrogen, and oxygen is burned in an excess of dioxygen, producing 9.191 g CO2 and 3.762 g H2O.  What percent by mass of oxygen is contained in the original sample?

76.  A) 54 %

77.  B) 96 %

78.  C) 73 %

79.  D) 43 %

80.  E) 939 %

 

 

ANS:      A             PTS:       1              DIF:        moderate            REF:       3.4

OBJ:       Calculate the percentage of C, H, and O from combustion data.

TOP:      stoichiometry | determining chemical formulas

 

76.  A 2.841 g sample of a hydrocarbon is burned in an excess of dioxygen, producing 7.794 g CO2 and water.  What mass of hydrogen is contained in the original sample?

77.  A) 7140 g

78.  B) 953 g

79.  C) 64 g

80.  D) 826 g

81.  E) 421 g

 

 

ANS:      A             PTS:       1              DIF:        moderate            REF:       3.4

OBJ:       Calculate the percentage of C, H, and O from combustion data. (Example 3.9)

TOP:      stoichiometry | determining chemical formulas

 

77.  A 2.445 g sample of a hydrocarbon is burned in an excess of dioxygen, producing 6.708 g CO2 and 5.492 g H2O.  What is the empirical formula of the hydrocarbon?

78.  A) CH4

79.  B) CH2

80.  C) C2H3

81.  D) CH3

82.  E) CH

 

 

ANS:      A             PTS:       1              DIF:        moderate            REF:       3.5

OBJ:       Determine the empirical formula of a binary compound.

TOP:      stoichiometry | determining chemical formulas

 

78.  A 0.4647-g sample of a compound known to contain only carbon, hydrogen, and oxygen was burned in dioxygen to yield 0.01962 mol of CO2 and 0.01961 mol of H2O. What is the empirical formula of the compound?

79.  A) CHO

80.  B) C3H3O2

81.  C) C2H2O

82.  D) C3H6O2

83.  E) C6H3O2

 

 

ANS:      D             PTS:       1              DIF:        moderate            REF:       3.4

OBJ:       Calculate the percentage of C, H, and O from combustion data. (Example 3.9)

TOP:      stoichiometry | determining chemical formulas

 

79.  A sample containing only carbon, hydrogen, phosphorus, and oxygen is subjected to elemental analysis.  After complete combustion, a 0.4946-g sample of the compound yields 0.7092 g of CO2, 0.4355 g of H2O, and 0.3812 g of P4O10.  What is the empirical formula of the compound?

80.  A) CH3PO

81.  B) C2H3PO

82.  C) C2H6P2O4

83.  D) C3H9PO

84.  E) CH2P4O13

 

 

ANS:      D             PTS:       1              DIF:        difficult REF:       3.4

OBJ:       Calculate the percentage of C, H, and O from combustion data. (Example 3.9)

TOP:      stoichiometry | determining chemical formulas KEY:       elemental analysis

MSC:     general chemistry

 

80.  A sample containing only carbon, hydrogen, and silicon is subjected to elemental analysis.  After complete combustion, a 0.1099-g sample of the compound yields 0.2193 g of CO2, 0.1346 g of H2O, and 0.07485 g of SiO2.  What is the empirical formula of the compound?

81.  A) CH3Si

82.  B) C2H4Si

83.  C) C4H12Si

84.  D) C6H12Si2

85.  E) CH2Si

 

 

ANS:      C             PTS:       1              DIF:        difficult REF:       3.4

OBJ:       Calculate the percentage of C, H, and O from combustion data. (Example 3.9)

TOP:      stoichiometry | determining chemical formulas KEY:       elemental analysis

MSC:     general chemistry

 

81.  Of the following, the only empirical formula is

82.  A)

83.  B)

84.  C)

85.  D)

86.  E)

 

 

ANS:      C             PTS:       1              DIF:        easy       REF:       3.5

OBJ:       Define empirical formula.

TOP:      stoichiometry | determining chemical formulas KEY:       empirical formula

MSC:     general chemistry

 

82.  Which of the following is the empirical formula for the molecule below?

 

1.    A) CHO

2.    B) CH3COOH

3.    C) C2H4O2

4.    D) CH2O

5.    E) none of the above.

 

 

ANS:      D             PTS:       1              DIF:        easy       REF:       3.5

OBJ:       Define empirical formula.

TOP:      stoichiometry | determining chemical formulas KEY:       empirical formula

MSC:     general chemistry

 

83.  Analysis of a compound showed that it contained 76.0 % fluorine atoms and 24.0 % carbon atoms by mass. What is its empirical formula?

84.  A) CF2

85.  B) C2F3

86.  C) CF3

87.  D) C2F5

88.  E) CF

 

 

ANS:      A             PTS:       1              DIF:        moderate            REF:       3.5

OBJ:       Determine the empirical formula of a binary compound from the masses of its elements. (Example 3.10)

TOP:      stoichiometry | determining chemical formulas KEY:       empirical formula

MSC:     general chemistry

 

84.  A sample of an oxide of antimony (Sb) contained 39.5 g of antimony combined with 13.0 g of oxygen. What is the simplest formula for the oxide?

85.  A) SbO2

86.  B) SbO

87.  C) Sb2O3

88.  D) Sb2O

89.  E) Sb2O5

 

 

ANS:      E              PTS:       1              DIF:        moderate            REF:       3.5

OBJ:       Determine the empirical formula of a binary compound from the masses of its elements. (Example 3.10)

TOP:      stoichiometry | determining chemical formulas KEY:       empirical formula

MSC:     general chemistry

 

85.  Chlorine was passed over 1.30 g of heated titanium, and 4.20 g of a chloride-containing compound of Ti was obtained. What is the empirical formula of the chloride-containing compound?

86.  A) TiCl2

87.  B) TiCl4

88.  C) TiCl

89.  D) TiCl3

90.  E) Ti2Cl3

 

 

ANS:      D             PTS:       1              DIF:        moderate            REF:       3.5

OBJ:       Determine the empirical formula of a binary compound from the masses of its elements. (Example 3.10)

TOP:      stoichiometry | determining chemical formulas KEY:       empirical formula

MSC:     general chemistry

 

86.  A 2.39-g sample of an oxide of chromium contains 1.48 g of chromium. Calculate the simplest formula for the compound.

87.  A) CrO5

88.  B) Cr2O

89.  C) CrO2

90.  D) CrO

91.  E) Cr2O3

 

 

ANS:      C             PTS:       1              DIF:        moderate            REF:       3.5

OBJ:       Determine the empirical formula of a binary compound from the masses of its elements. (Example 3.10)

TOP:      stoichiometry | determining chemical formulas KEY:       empirical formula

MSC:     general chemistry

 

87.  A compound is composed of only C and H. It contains 92.26 % C. What is its empirical formula?

88.  A) C2H5

89.  B) C2H3

90.  C) C3H4

91.  D) CH

92.  E) CH2

 

 

ANS:      D             PTS:       1              DIF:        easy       REF:       3.5

OBJ:       Determine the empirical formula from the percentage composition. (Example 3.11)

TOP:      stoichiometry | determining chemical formulas KEY:       empirical formula

MSC:     general chemistry

 

88.  A compound composed of only C and F contains 17.39 % C by mass.  What is its empirical formula?

89.  A) CF3

90.  B) CF

91.  C) C2F

92.  D) CF4

93.  E) CF2

 

 

ANS:      A             PTS:       1              DIF:        easy       REF:       3.5

OBJ:       Determine the empirical formula from the percentage composition. (Example 3.11)

TOP:      stoichiometry | determining chemical formulas KEY:       empirical formula

MSC:     general chemistry

 

89.  A hydrocarbon, subjected to elemental analysis, was found to contain 85.63 % carbon and 14.37 % hydrogen by mass.  What is the empirical formula of the hydrocarbon?

90.  A) CH4

91.  B) C2H4

92.  C) C6H

93.  D) C10H

94.  E) CH2

 

 

ANS:      E              PTS:       1              DIF:        easy       REF:       3.5

OBJ:       Determine the empirical formula from the percentage composition. (Example 3.11)

TOP:      stoichiometry | determining chemical formulas KEY:       empirical formula

MSC:     general chemistry

 

90.  A particular compound contains, by mass, 41.4 % carbon, 3.47 % hydrogen, and 55.1 % oxygen.  A 0.050-mol sample of this compound weighs 5.80 g.  The molecular formula of this compound is

91.  A)

92.  B)

93.  C)

94.  D)

95.  E)

 

 

ANS:      D             PTS:       1              DIF:        moderate            REF:       3.5

OBJ:       Determine the empirical formula from the percentage composition. (Example 3.11)

TOP:      stoichiometry | determining chemical formulas

KEY:       molecular formula                           MSC:     general chemistry

 

91.  What is the empirical formula of an oxide of nitrogen that contains 25.93 % nitrogen by mass?

92.  A) NO2

93.  B) N2O

94.  C) N2O3

95.  D) NO

96.  E) N2O5

 

 

ANS:      E              PTS:       1              DIF:        moderate            REF:       3.5

OBJ:       Determine the empirical formula from the percentage composition. (Example 3.11)

TOP:      stoichiometry | determining chemical formulas KEY:       empirical formula

MSC:     general chemistry

 

92.  The analysis of an organic compound showed that it contained 1.386 mol of C, 0.0660 mol of H, 0.924 mol of O, and 0.462 mol of N. How many nitrogen atoms are there in the empirical formula for this compound?

93.  A) 9

94.  B) 7

95.  C) 2

96.  D) 4

97.  E) 3

 

 

ANS:      B             PTS:       1              DIF:        easy       REF:       3.5

OBJ:       Determine the empirical formula from the percentage composition. (Example 3.11)

TOP:      stoichiometry | determining chemical formulas KEY:       empirical formula

MSC:     general chemistry

 

93.  A sample containing 0.700 mol of a compound is composed of 4.21 ´ 1023 atoms of sodium, 24.79 g of chlorine atoms, and 33.57 g of oxygen atoms. The formula of the compound is

94.  A)

95.  B)

96.  C)

97.  D)

98.  E)

 

 

ANS:      A             PTS:       1              DIF:        moderate            REF:       3.5

OBJ:       Determine the empirical formula from the percentage composition. (Example 3.11)

TOP:      stoichiometry | determining chemical formulas KEY:       empirical formula

MSC:     general chemistry

 

94.  The analysis of an organic compound showed that it contained 0.0700 mol of C, 0.175 mol of H, and 0.0350 mol of N. Its molecular mass is 86 amu. How many atoms of carbon are there in the empirical formula for the compound, and how many are in the molecular formula?

95.  A) empirical = 2, molecular = 3

96.  B) empirical = 2, molecular = 6

97.  C) empirical = 2, molecular = 4

98.  D) empirical = 5, molecular = 10

99.  E) empirical = 3, molecular = 3

 

 

ANS:      C             PTS:       1              DIF:        moderate            REF:       3.5

OBJ:       Determine the empirical formula from the percentage composition. (Example 3.11)

TOP:      stoichiometry | determining chemical formulas

KEY:       molecular formula                           MSC:     general chemistry