Solutions for Some Chapter 3 practice problems(CHEM 1411)
1.
a) C5H10O2 + O2 → CO2 + H2O
(2, 13, 10, 10)
b) PCl5 + H2O → H3PO4 + HCl
(1, 4, 1, 5)
c) Al(OH)3 + H2SO4 → Al2(SO4)3 + H2O
(2, 3, 1, 6)
d) Na + H2O → NaOH + H2
(2, 2, 2, 1)
2. C7H16 + O2 → CO2 + H2O
SO3 + H2O → H2SO4
Li + H2O → LiOH + H2
ZnCO3 → ZnO + CO2
(1, 11, 7, 8)
(1, 1, 1)
(2, 2, 2, 1)
(1, 1, 1)
3. Determine formula weight:
Ca(C2H3O2)2 : 40g + (4x12) + (6x1) + (4x16) = 158g
4. Percentage compositions:
a) benzoic acid, C7H6O2 molar wt. is:
(7x12) + (6x1) + (2x16) = 84 + 6 + 32 = 122g
%comp. C=> (7*12/122) x 100 = 68.85%, H => (6*1/122) x 100 = 4.92%
O  (2*16/122) x 100 = 26.23%
5. Let x be the fractional abundance of isotope 70.926-Ga. Then:
69.726 = 70.926x + 68.926(1-x)  x = 0.8000/2 = 0.4000
% abundance of isotope 70.926 is 0.4000(100) = 40%
% abundance of isotope 68.926 is 60%.
6. (a) 0.005g │ 1mole H2O
│ 18gH2O
= 2.77 x 10-4 mole = 3 x 10-4 mole
(b) 3.5moles │6.022 x 1023 molecules│3atoms = 6.32 x 1024 atoms
│ 1 mole
│1molecule
(c) 2.5 moles │ 32g O2 = 80g O2
│ 1 mole
(d) F.W. C6H12O6 = 180g = glucose = glu
(10.5g-glu)│1mole-glu│6.02 x 1023molecules-glu│6C = 2.1 x 1023 C atoms
│180g-glu) │1 mole-glu
│1 molecule-glu)
7. (a) CWHXNYOZ Epirical Formula? 62.19% C, 5.21% H, 12.1% N, 20.7% O
Assume 100g sample. Then:
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62.1g/12g = 5.175moles C; 5.21g/1g = 5.21moles H; 12.1g/14g = 0.86 mole N
20.7g/16g = 1.29 moles O
Divide smallest mole number into each calculated mole:
5.175/0.86 = 6; 5.21/0.86 = 6; 0.86/0.86 = 1; 1.29/0.86 = 1.5
multiply all subscripts by 2 to get all of them as whole numbers.
Therefore: Empirical Formula of CWHXNYOZ is C12H12N2O3
(b) KXCYOZ Empirical Formula?
Divide smallest mole number into each mole.
i.e. (0.104/.052) = 2moles K; (0.052/0.052) = 1mole C; (0.152/0.052) = 3moles O ; Therefore K2CO3
(c) Get equivalent mole value for each mass.
Fe 11.66g/56g = 0.21 mole; O 5.01g/16 = 0.313 mole
Divide smallest mole into each calculated mole.
Fe  1, O  1.5 , Multiply subscripts by 2  Therefore Empirical formula For FexOy is Fe2O3
8. (a) First find empirical formula mass.
If molecular formula mass is 90g 
HCO2  1 + 12 + 2(16) = 45g
90g/45g = 2  (HCO2)2 = H2C2O4
(b) % composition given, Molar mass given  206g for CxHyOz
Moles of C is 75.69g/12g = 6.31moles
Moles of H is 8.8g/1g = 8.8moles
Moles of O is 15.51g/16g = 0.97mole
C  6.31/0.97 = 6.5; H  8.8/0.97 = 9.07; O  0.97/0.97 = 1.00
Multiply by 2. Then C  13; H  18; O  2
Then CxHyOz is C13H18O2 = empirical and molecular formula. OR
mole of C = (206g/12g)( 0.7569) = 12.99 mole  13 moles = x
mole of H = (206g/1g)( 0.088) = 18.12 mole  18 moles = y
mole of O = (206g/16g)( 0.1551) = 1.99 moles  2 moles = z
 C13H18O2 = molecular formula.
9. (3/11)5.86g = 1.60g C; (1/9)1.37g = 0.15g H  toluene sample mass is 1.75g.
% mass C is 1.60g/1.75g x 100 =91.43%
% mass H is 0.15/1.75 x 100 = 8.6%
So moles C in 92g/mol is (0.9143 x 92g)/12g = 7.01 moles  7.00 moles = x
So moles H in 92g/mol is (0.086 x 92g)/1g = 7.9 moles  8.00 moles = y
So for CxHy molecular formula is C7H8
2
10. (3/11)(0.2829g) = 0.07715g C; (1/9)(0.1159g) = 0.01288g H.
Mass of C & H in sample is (0.07715 + 0.01288) = 0.09003g
Therefore sample contains O of mass (0.1005- 0.09003) = 0.01047g
So the compound is CxHyOz
Moles of C in molecular mass is (0.07715g x 156g)/(0.1005 x 12g) = 9.97  10
Moles of H in molar mass is (0.01288g x 156g)/(0.1005g x 1g) = 19.99  20
Moles of O in molar mass is (0.01047g x 156g)/(0.1005g x 16g) = 1.01  1.00
Then x = 10; y = 20; z = 1
Therefore molecular formula of CxHyOz is C10H20O
11. (a) 16 CO2/2 C8H18 = X/1.5; X = 12.0 moles CO2
(b) 18H2O/2 C8H18 = X/1.5; X = 13.5 moles H2O  13.5x18g/mol = 243g H2O
(c) (90g water)/18g/mol = 5 moles; (25mol)/(18mol) = (Xmoles O2)/5moles H2O; X = 6.94
moles oxygen gas.
12. 4NH3 + 5O2  4NO + 6H2O
(a) 2.50g NH3 is 0.147mole  0.147 mole NO produced
2.85g O2 is 0.089mole  (4/5)(0.089mole) = 0.071mole NO produced.
Therefore O2 is limiting reagent.
(b) Grams NO formed is (0.071mole x 30g) = 2.13g NO
(c) NH3 is excess reactant. How much? 0.147mole – (4/5)(0.089mole)= 0.147 – 0.071= 0.076 mole =
17g(0.076mol) = 1.29 grams excess NH3.
13. C6H6 + Br2 → C6H5Br + HBr (30g benzene, 65g bromine gas)
C6H6 mol wt. is 78g; Br2 mol wt. is 160g; C6H5Br mol wt. is 157g
All stoichiometric relationships are 1 to 1. Limiting reagent?
For Benzene (157g/78g) = (X/30g); X = 60.38g bromobemzene produced.
For Bromine gas (157g/160g) = (X/65g); X = 63.78g bromobenzene produced.
Therefore benzene is limiting reagent. Then theoretically, 30g benzene should
produce 60.38g of bromobenzene for a 100% yield. But since only 56.7g was
produced, percent yield is (56.7g/60.38g) x 100 = 94% yield.
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