Chemistry 112
First Hour Exam
Please show all work for partial credit
1. (10 points) Perform the following unit conversions:
34.56 km to nm
34.56km ×
1000 m
1nm
×
= 3.456 x1010 nm
−9
1km
1x10 m
3.5 tons to kg
35
. ton ×
2000lbs
1kg
×
= 3,175kg
1ton
2.2046lbs
1.0 g/cm3 to lbs/gal
1kg
2.2046lbs 1cm3 1000ml
1l
4qt
1.0 cm3 ×
×
×
×
×
×
= 8.345lbs gal
1000g
1kg
1ml
l
10567
.
qt gal
g
1.0 g/cm3 to lbs/ft3
3
1kg
2.2046lbs  2.54cm 12in 
1.0 cm3 ×
×
×
×
 = 62.43 lbs 3
ft
1000 g
1kg
 1in
ft 
g
2. (10 points) Perform the following mathematical operations and express the result to the
correct number of significant figures:
4.184 x 1006.2 x (25.27-24.16) =
4.184 4 sig fig, 1006.2 5 sig fig, 25.27-24.16 = 1.11 3 sig fig
4.67x103 or 4670
9.2× .00065
=
8.321 − 8.315
9.2 2 sig fig, .00065 2 sig fig, 8.321-8.315=.006 = 1 sig fig
.9967 that rounds to 1
.001956 x .190056 x .195600 = .001956 4 sig fig, .190056 6 sig fig, .195600 6
7.271x10-5
3. (10 points) Name the following elements
Any three transition metals __Fe, Co, Ni, Cu, etc_
A halide in the second period __________F___________
An alkaline earth metal in the fourth period _______Cu__________
2
4. (10 points) Give the names of the following compounds
CaS _Calcium sulfide
Li3N _Lithium nitride
HNO3 (aq) _Nitric Acid
N2H4 __Dinitrogen tetrahydride
CrO3 __Chromium(VI) oxide
5. (10 points) Give the molecular formula of the following compounds
Lead (IV) oxide __PbO2 __________
Magnesium phosphate __Mg3(PO4)3_______
Sulfur tetrafluoride __SF4____________
Sulfuric Acid ____H2SO4(aq) _________
Tetraphosphorous hexoxide ___P4O6______
6. (10 points) What is the % composition of each element in Na 3PO4
3 Na for 3x22.99 = 68.97
1 P for 1x30.97 = 30.97
4 O for 4x16.0 = 64.00
Total = 163.94
% Na = 68.97/163.94 x 100% = 42.07%
% P = 30.97/163.94 x 100% = 18.89%
% O = 64/163.94 x 100% = 39.04%
For a quick check all %’s add up to 100%
3
7. (10 points) A compound contains 47.08% C, 6.59% H and 46.33% Cl by mass; the molar
mass of the compound is 153g/mol. What are the empirical and molecular formulas of the
compound?
g C = 153x.4708 = 72.03g
g H = 153 x .0659 = 10.08 g H
g Cl = 153 x .4633 = 70.88 g Cl
Divide by molar mass of atom
C 71.03/12 ~ 6
H 10.08 /1 ~10
Cl 70.88/ 35.44 ~ 2
Molecular formula = C 6H10Cl2
Empirical Formula = C 3H5Cl
8. (10 points) Balance the following chemical equations:
Na2SiF6(s) + 4Na(s) 6 Si(s) + 6NaF(s)
2 Na on the left hand side so DON”T start with Na!
Si
F
Na
1
6
2 4x1
1
1x6
6
2NaCl(aq) + 2H2O(l) 6 Cl2(g) + H2(g) + 2NaOH(aq)
Cl 1x2
Na 1x2
O
H
2
1x2
2x2
2
1x2
1x2
2
4
9. (10 points) When silver is heated in the presence of sulfur, it will form silver sulfide in the
reaction:
16 Ag(s) + S 8(s) 6 8Ag2S(s)
If I started with reaction with 5g of silver
A. How much Sulfur (S 8) would I need for this reaction?
5g Ag ×
1 mol Ag 1 mol S 8 256.5g S 8
×
×
= .753g
107.9g
1 mol Ag
mol S 8
B. How much silver sulfide would be formed?
5g Ag ×
1 mol Ag 8 mol Ag 2 S 247.87g Ag 2S
×
×
= 574
. g
107.9g
1 mol Ag
mol Ag 2 S
Or the quick and dirty answer, mass of product = mass of reactants = 5 + .753 = 5.753
10. (10 points) I am going to burn aluminum in bromine gas using the following reaction:
2Al(s) + 3Br2(g) 62AlBr3(s)
If I start with 6 g of aluminum and 44 g of Bromine gas, and my yield of aluminum
bromide is 45.0 g, what is the % yield for my reaction?
Assume Al is limiting:
6g Al ×
1 mol Al
2 mol AlBr3 266.8g AlBr3
×
×
= 59.31g
26.98g Al
2 mol Al
1mol AlBr3
Assume Br2 is limiting:
44 g Br2 ×
1mol Br2
2mol AlBr3 266.68g AlBr3
×
×
= 48.95g
159 .8g Br2
3mol Br2
mol AlBr3
Br2 is lower so limiting reagent,
% Yield = 45/48.95 x 100% = 91.93%