Chem 1405 Solution to homework chap4
1. Calculate the formula mass of aspirin, C9H8O4. (Section 4.1)
a. 29.02 amu
b. 180.15 amu c. 21 amu
2. Which of the following statements best describes a mole? (Section 4.2)
a. Mole is an abbreviation for the word molecule.
b. A mole is a specific type of molecule.
c. A mole is a unit of measurement much like a dozen, only rather than having 12 units, 1 mole has
6.022 × 1023 units.
3. Calculate the molar mass of aspirin, C9H8O4. (Section 4.2)
a. 29.02 g/mol b. 18.15g/mol c. 180.15 g/mol
Hint: For any molecule, molecular mass (amu) = molar mass (grams). The unit of molar mass is g/mol,
but in your book you may see that sometime only gram is used as the unit.
4. How many moles of ibuprofen C13H18O2 does a 500 mg tablet contain and how many C13H18O2
molecules does a tablet contain? (Section 4.2)
a. 2.42 ×10-3 moles, 1.44x1021 b. 103 moles, 1.44x1023 c. 2.42, 1.44x1021
d. 2.42 ×10-3 moles, 1.44x1023
Hint: it’s very important to remember the relationship between mass and number of moles, n=m/M, the
molar mass of C13H18O2 is 13x12+1x18+16x2=206g/mole, n=0.5g/(206g/mole)=0.00242 mole
The number of molecules #=nNA, here NA = Avogadro’s number=6.022 x 1023
number of molecules #=nNA=0.00242molex6.022 x 1023=1.44x1021
5. How many total hydrogen atoms are contained in a 500 mg tablet of ibuprofen (C13H18O2) contain?
(Section 4.3)
a. 1.12 × 1027 hydrogen atoms
b. 2.63 × 1022 hydrogen atoms
c. 4.36 × 10-2 hydrogen atoms
Hint : it’s important to remember: the number of atoms or molecules #=nNA, here NA = Avogadro’s
number=6.022 x 1023. From previous question, you know there are 2.42 ×10-3 moles moles in 500 mg
tablet. And in each C13H18O2 molecule there are 18 hydrogen atoms.
So the number of hydrogen atoms=18x2.42 ×10-3x6.022 x 1023=2.63 × 1022
6. A sample contains 44.2 % Fe and 55.8% Cl. What is the empirical formula?
(Answer: FeCl2)
Hint:In order to calculate empirical formula , we must know moles.
However, in order to calculate moles, we need gram amounts – and we only have %.
Therefore, assume a sample size: 100.0 g is the easiest sample size to assume.
Now we can use molar mass to calculate moles.
From the periodic chart we find the molar masses:
Fe = 56.9g/mol; Cl = 35.5 g/mol.
7. A compound containing only the elements C, H, and O is known to have a molar mass of 88 ± 1 g. The
compound is known to contain 54.6%C and 9.0% H by mass. What is the molecular formula for this
compound?
8. What are the coefficients when the following equation is balanced?
PCl3 (l) + H2O (l) H3PO3 (aq) + HCl (aq)
a. 1,1,2,2, b. 1,2,1,2 c. 1,3,1,3 d. 3,1,3,1
9. What are the coefficients when the following equation is balanced?
KClO3 (s)KCl (s) + O2 (g)
a. 1,1,1 b. 2,1,2 c. 3,2,1 d. 2,2,3
10. What is the oxidation state of phosphorous in phosphorous trichloride, PCl3? (Section 4.7)
a. +1
b. +3 c. -3
11. Which of the following underlined atoms contains the oxidation number as -1?
(a) Cs2O
(b) CaC2
(c) SO42-
(d) PtCl42-
(e) Na2O
12. Which of the following represents a half-reaction for the oxidation of magnesium? (Section 4.7)
a. Mg + 2e Mg2+
b. Mg + 2e Mg2-
c. Mg Mg2+ + 2e-
13. Which of the following represents the correct chemical equation for the combustion of propane,
C3H8? (Section 4.7)
a. C3H8 + 5O2  3CO2 + 4H2O
b. C3H8 + O2 4H2 + 3C + O2
c. C3H8 + 7O  3CO2 + H2O + 3H2
14. How many moles of oxygen are required to react with 5 moles of hydrogen to produce water
according to the reaction 2H2 + O2  2H2O? (Section 4.8)
a. 0.4 moles oxygen
b. 5 moles oxygen
c. 10 moles oxygen
d. 2.5 moles oxygen
Hint: the unit-conversion factor between oxygen and hydrogen is (1 mole O2/2 mole H2)
Mole of O2 required= 5 mole H2xunit-conversion factor= 5 mole H2x(1 mole O2/2 mole H2)=2.5 mole
15. How many grams of Al(OH)3 are required to neutralized 10.0 g of stomach acid, HCl, according to
the equation Al(OH)3 + 3HCl  3H2O + AlCl3? (Section 4.8)
a. 3.33 g
b. 21.4 g
c. 7.10 g
hint: For stoichiometry using chemical equation, always use moles for calculation.
Al(OH)3 + 3HCl  3H2O + AlCl3
n=m/M
X (1 Al(OH)3/3 HCl)
m=nXM
10.0 g HCl-------------------->0.274mol HCl----------------------->0.091mol Al(OH)3 ---------------------->7.10g
10.0g/(36.45g/mol)
0.274x(1/3)
0.091molx(77.98g/mol)