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Exam #5 Review
SHOW ALL WORK ON A SEPARATE PIECE OF PAPER!!
Determine the Molar Mass of the following compounds.
1. (NH4)3PO4
N: 3 x 14.01 = 42.03
H: 12 x 1.008 = 12.096
P: 1 x 30.97 = 30.97
ADD ALL THE MASSES
TOGETHER!
O: 4 x 16.00 = 64.00
149.096 g
2. O2Cl4
O: 2 x 16.00 = 32.00
Cl: 4 x 35.45 = 141.80
ADD ALL THE MASSES
TOGETHER!
173.80 g
3. CuClO3
Cu: 1 x 63.55 = 63.55
Cl: 1 x 35.45 = 35.45
O: 3 x 16.00 = 48.00
ADD ALL THE MASSES
TOGETHER!
147.00 g
4. Mg(OH)2
Mg: 1 x 24.30 = 24.30
O: 2 x 16.00 = 32.00
H: 2 x 1.008 = 2.016
58.316 g
ADD ALL THE MASSES
TOGETHER!
Determine % composition of the following elements.
First, find the total molar mass (g) of the compound.
Then use the formula!
% comp =
total mass of ELEMENT __ x 100
MOLAR MASS OF COMPOUND
5. What % of chlorine is found in O2Cl4?
O2Cl4
O: 2 x 16.00 = 32.00
Cl: 4 x 35.45 = 141.80
Total Molar Mass = 173.80 g
141.80 X 100 = 81.59 %
173.80
Divide & multiple by 100 to get a percent!
6. What % of chlorine is found in CuClO3?
CuClO3
Cu: 1 x 63.55 = 63.55
Cl: 1 x 35.45 = 35.45
O: 3 x 16.00 = 48.00
35.45 X 100 = 24.12 %
147.00
Divide & multiple by 100 to get a percent!
Total Molar Mass = 147.00 g
7. What % of oxygen is found in Mg(OH)2?
Mg(OH)2
Mg: 1 x 24.30 = 24.30
O: 2 x 16.00 = 32.00
32.00 X 100 = 54.87 %
58.316
H: 2 x 1.008 = 2.016
Divide & multiple by 100 to get a percent!
Total Molar Mass = 58.316 g
Solve the following to find the empirical formula.
8. What is the empirical formula for a compound that consists of 74.47% C, 8.75% H and 17.36% N?
a. Change percent to mass (assume 100 g)
74.47% C = 74.47 g C
8.75% H = 8.75 g H
17.36% N = 17.36 g N
b. Mass to Moles (Divide the masses by their mass on the periodic table! This will equal MOLs)
74.47 g C
1 mol C
= 6.20 mol C
12.01 g C
8.75 g H
1 mol H
= 8.68 mol H
1.008 g H
17.36 g N
1 mol N
= 1.24 mol N
14.01 g N
c. Divide by Small (Divide by the smallest mol value)
C: 6.2 mol / 1.24 = 5
H: 8.68 mol / 1.24 = 7
N: 1.24 mol / 1.24 = 1
d. Times till whole! Multiply until you get a whole number! (since all are whole numbers, you
can skip step 4!)
Final Answer/Empirical Formula: C5H7N
9. For these below, just reduce!
a. C6H12O6
(Divide by 6)
b. N3Br27
(Divide by 3)
CH2O
NBr9
c. O15F24
(Divide by 3)
O5F8
Mole Conversions
10. 1 mol is equal to molar mass. These units are in grams (g).
11. 1 mol is equal to 6.02 x 1023 particles. This number is known as Avogadro’s Number!
12. What are the 3 types of particles?
atoms, ions and molecules
13. How many molecules are found in 2.94 moles of CO2?
2.94 mol of CO2
6.02 x 10
23
1 mol CO2
molecules of CO2
CONVERSION FACTORS
1 mol = 6.02 x 10 23 molecules
1 mol = molar mass (g)
= 1.77 x 1024 molecules of CO2
14. How many moles are found in 79.2 grams of Fe?
79.2 g of Fe
1 mol Fe
= 1.42 mol Fe
55.85 g Fe
15. What is the mass (grams) of 4.16 moles of Ca3(PO4)2?
4.16 mol of Ca3(PO4)2
310.18 g Ca3(PO4)2
Molar Mass of Ca3(PO4)2
Ca: 3 x 40.08 = 120.24
P: 2 x 30.97 = 61.94
O: 8 x 16.00 = 128.00
310.18 g
= 1290.35 g Ca3(PO4)2
1 mol Ca3(PO4)2
16. How many moles contain 5.92 x 1024 atoms of P?
5.92 x 1024 atoms of P
1 mol of P
Make sure to cancel out your units!
= 9.83 mol of P
6.02 x 1023 atoms of P
Molar Mass of NaNO3
17. How many molecules are in 74.8 grams of NaNO3?
74.8 g of NaNO3
1 mol of NaNO3
85.00 g NaNO3
6.02 x 1023
molecules of NaNO3
1 mol of NaCl
23
6.02 x 10
molecules of NaCl
= 5.30 x 1023 molecules NaNO3
1 mol of NaNO3
18. What is the mass (grams) of 3.71 x 1025 molecules of NaCl?
3.71 x 1025
molecules of NaCl
Na: 1 x 22.99 = 22.99
N: 1 x 14.01 = 14.01
O: 3 x 16.00 = 48.00
85.00 g
58.44 g NaCl
1 mol of NaCl
Molar Mass of NaCl
Na: 1 x 22.99 = 22.99
Cl: 1 x 35.45 = 35.45
58.44 g
= 3601.53 g NaCl