Chem 1151 Exam 1B-1 (46 pts) February 14, 2007
Name______KEY______GREEN______Lab Sec_____
SHOW ALL WORK. N = 6.022E23
1. a. (4) During your study abroad semester in South Africa, you discover a shiny piece of metal and you wonder if it would pay for your tuition next year. Back in the lab, you found that it weighed 106.48 g. After filling a 25.00-mL graduated cylinder with water up to the 15.24 mL mark, you carefully add the piece of metal. Calculate density of your find if the water level rose to 20.29 mL. Assume that the metal is pure.
d = m/V = 106.48g/(20.29-15.24) = 10..48/5.05 = 21.09 g/cm3
b. (1) You found these density values on the web. What is the identity of the metal? The metal is Pt
metal / Ag / Au / PtDensity (g/cm3) / 10.49 / 19.3 / 21.09
Price per ounce / $13.93 / $672.30 / $1203.
c. (3) How much is your metal worth? 1 pound = 453.6 g = 16 oz.
106.48g/453.6 g/lb [16 oz/lb] $1203/oz = $4518
2 (3) Express your answer to this calculation using the correct number of sig. figs.
(124.37 – 67.578)/16.284 = 56.792/16.284 = 3.48759 = 3.488
3. (3) Which has the largest volume? 75. cm3 or 5.0 in3.
1 in = 2.54 cm.
5.0 in3 (2.54cm/in)3 = 81.93 = 82 in3 > 75. cm3
4. You were given four bottles marked A, B, C and D and told that each bottle contained a compound comprised only of hydrogen and nitrogen. In addition, the bottles were labeled with the mass of hydrogen that combines with 1.00 g of nitrogen.
a. (2) Your first assignment is to find the two bottles containing the same compound.
A and C contain the same compound (Law of Definite Proportions)
Bottle / A / B / C / DMass of H combining with 1.00 g N / 0.0240 g / 0.144 g / 0.0240 g / 0.216 g
b. (4) Your second assignment is to determine the empirical formula for Sample D.
0.216 g H/1.01 g/mol = 0.21386 mol H
1.00 g N/14.00 g/mol = 0.0714 mol N
Divide by 0.0714 to get 3 H to 1 N or EF = NH3
c. (3) Finally, do the data for Samples A and B confirm the Law of Multiple Proportions? Show your work.
Sample B = (0.144g H/1.01g ) / (0.024 g N/14.00 g/mol) = 6/1
Sample A
H and N combine to form more than one cmp and the ratio of the mass of H that combines with 1.00 g N is a ratio of small whole numbers: 6 or 1.
5. The combustion of ethane is described by
2C2H6(g) + 7O2(g) à 6H2O(ℓ) + 4C EF = NH3 (g)
a. (2) What is the empirical formula of ethane? EF = CH3
b. (3) How many moles of oxygen are needed to completely burn 3.42 mol ethane? Let ethane = E
3.42 mol E (7 mol O2/2 mol E) = 12.0 mol O2
c. (3) How many grams of water will be produced when 3.42 mol ethane is burned in excess oxygen?
3.42 mol E (6 mol water/2 mol E) 18.02 g/mol = 185 g water
d. (4) If 500.0 g ethane and 1500.0 g oxygen are reacted, how many mols of CO2(g) will be produced?
Molar mass of ethane = 30.08 g/mol and molar mass of O2 = 32.00 g/mol
500.0 g E/30.08 g/mol = 16.62 mol E
16.62 mol E (7 mol oxygen/2 mol E) = 58.2 mol oxygen needed
1500.0/32 g/mol = 46.9 mol oxygen < 58.2 mol oxygen so O2 = LR
1500.0 g O2/ 32.00 g/mol (4 mol CO2/ 7 mol O2) = 26.8 mol CO2.
6. 5. (4) How many atoms of oxygen are present in 100.0 g of Cr2(CO3)3? Molar mass of this compound is 284.03 g/mol.
100.0 g/284.03 g/mol (9 mol O/1 mol volec) 6.022E23 = 1.908E24 O atoms
7. Now consider the combustion of ammonia.
4NH3(g) + 5O2(g) à 6H2O(ℓ) + 4NO(g)
a. (3) Determine the % composition by mass of nitrogen and hydrogen in ammonia.
NH3 has molar mass 14.00 = 3 x 1.01 = 17.03 g/mol
N % mass = (14.00/17.03) x 100 = 82.2 % N
H% = 100.0% -82.2% = 17.8% H
b. (4) When 170.3 g ammonia burned in excess oxygen, 250.0 g NO were produced. What is the % yield of this reaction?
170.3g/17.03g/mol (4 mol NO/4 mol ammonia) 30.00 g/mol = 300.0 g NO = theoretical yield.
% yield = (actual/theoretical) x 100 = (250.0/300.0) 100
= 83.3% = % yield
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