1. Given 4.50 g of KClO3
a. How many moles of KClO3 is that?
b. How many grams of K are in the sample?
c. How many g of O are in the sample?
d. How many moles of KCl are in the sample?
e. How many KClO3 molecules is that?
f. How many moles of K are in the sample?
g. How many moles of O are in the sample?
h. What is the per cent, by mass, of oxygen in KClO3?
i. What should be the mass of one KClO3 molecule?
j. What should be the mass of 1.00 mole of KClO3 molecules?
2. Given the following data:
Mass of crucible + cover 27.56 g
Mass of crucible + cover + CuSO4·5H2O 34.16 g
Mass of crucible + cover + CuSO4 ______
Calculate what the final mass should be, and show all calculations. Show all work.
3. Given a carbon oxygen compound, CXOY, containing 24.0 grams of carbon and 64.0 grams of oxygen, what is the correct formula of the compound? Show all reasoning.
4. Given the following data for the decomposition of sodium sulfate hydrate (sodium sulfate is Na2SO4)
mass of crucible + cover 25.00 g
mass of crucible + cover + hydrate 30.00 g
mass of crucible + cover + anhydrate 27.20 g
a. Calculate the number of water molecules in each hydrate molecule
b. Write the balanced equation for the reaction
c. Write the mole ratio for the reaction
d. Write the gram ratio for the reaction
5. What is the cost of purchasing enough ore to supply 1.00 kg of K when mined from a site which is 13.5% KCl by mass, at a cost of $3.30 per metric ton?
6. Balance the equations:
Al + S8 ==> Al2S3
H 3PO4 + Mg(OH)2 ==> Mg3(PO4)2 + H2O
Rb + RbNO3 ==> Rb2O + N2
Solutions
1. Given 4.50 g of KClO3
a. How many moles of KClO3 is that? n=m/MW = 4.50g/122.6g/mole = .0367 moles of KClO3
b. How many grams of K are in the sample? in each KClO3 there is 1 K, so n= .0367 moles of K;
n = m/AW; .0367 moles = mass /(39.1 g/mole) = 1.43 g
c. How many g of O are in the sample? in each KClO3 there are 3 O's, so there are 3 (.0367) moles of O = .110 moles of O
since n=m/AW, .110moles = m/(16.0 g/mole), m = 1.76 g of O
d. How many moles of KCl are in the sample? in each KClO3 there in 1 KCl. In .0367 moles of KClO3 there are .0367 moles of KCl
e. How many KClO3 molecules is that? .0367 moles (6.022E23/mole) = 2.21 E22 KClO3 molecules
f. How many moles of K are in the sample? in each KClO3 there is 1 K, so n= .0367 moles of K
g. How many moles of O are in the sample? in each KClO3 there are 3 O's, so there are 3 (.0367) moles of O = .110 moles of O
h. What is the per cent, by mass, of oxygen in KClO3? % = (48.0/122.6) * 100 = 39.2 %
i. What should be the mass of one KClO3 molecule? 122.6 AMU
j. What should be the mass of 1.00 mole of KClO3 molecules? 122.6 g
2. Given the following data:
Mass of crucible + cover 27.56 g
Mass of crucible + cover + CuSO4·5H2O 34.16 g
Mass of crucible + cover + CuSO4 ______
Calculate what the final mass should be, and show all calculations. Show all work.
Mass of CuSO4·5H2O = 34.16 g - 27.56 g = 6.60 g
CuSO4·5H2O ==> CuSO4 + 5H2O theoretical masses 250 160 90 experimental masses 6.60 g x y 250/6.60 = 160/x x = 4.22g
250/6.60 = 90/y y = 2.38 gFinal mass = 27.56 g + 4.22 g = 31.78 g or
Final mass = 34.16 g - 2.38 g = 31.78 g
3. Given a carbon oxygen compound, CXOY, containing 24.0 grams of carbon and 64.0 grams of oxygen, what is the correct formula of the compound? Show all reasoning.
Calculate the number of moles of each type of atom, then calculate the simplest whole number ratio, then construct the formula. C O number of moles 24.0 g/(12.0 g/mole) = 2.0 moles 64.0 g/(16.0 g/mole) = 4.0 moles divide both by the smaller to determine the simplest whole number ratio:
2.0 moles / 2.0 moles = 1.0 4.0 moles / 2.0 moles = 2.0 simplest whole number ratio 1 2 formula is CO2
4. Given the following data for the decomposition of sodium sulfate hydrate (sodium sulfate is Na2SO4)
mass of crucible + cover 25.00 g
mass of crucible + cover + hydrate 30.00 g
mass of crucible + cover + anhydrate 27.20 g
a. Calculate the number of water molecules in each hydrate molecule
b. Write the balanced equation for the reaction
c. Write the mole ratio for the reaction
d. Write the gram ratio for the reaction
a. mass of anhydrate (Na2SO4) = 27.20 g -25.00 g = 2.20 g
mass of water (xH2O) = 30.00 g -27.20 g = 2.80 g
Na2SO4·xH2O ==> Na2SO4 + xH2O theoretical masses 142 + 18x 142 18x experimental masses 5.00 2.20 2.80 142/2.20 = 18x /2.80 x = 10.0 water molecules in each hydrate molecule
b. Na2SO4·10H2O ==> Na2SO4 + 10H2O
c. 1 : 1 : 10
d. 322 : 142 : 180
5. What is the cost of purchasing enough ore to supply 1.00 kg of K when mined from a site which is 13.5% KCl by mass, at a cost of $3.30 per metric ton?
a. Calculate the % of K in KCl. This gives you the ratio of mass of K to mass of KCl
The molecular weight of KCl is 74.6; 39.1 of the 74.6 is K, by mass.; The percent of K in KCl is 39.1 * 100 / 74.6 = 52.4%
But not really because the mixture is only 13.5% pure KCl. So calculate .135 (52.4%) = 7.07 % Kb. If 100 g of KCl contains 7.07 g of K, how many g of KCl contain 1000 g of K?
100/7.07 = x/1000 g
x = 14144 g or 1.41 E 4 g.c. Convert to metric tons
1.41E4 g (1 metric ton/1000000 g) = .0141 metric tons.
d. Calculate cost
Cost = .0141 metric tons ($3.30/Metric ton) = $ 0.0465
6. Balance the equations:
16Al + 3S8 ==> 8Al2S3
2H3PO4 + 3Mg(OH)2 ==> Mg3(PO4)2 + 6H2O
10Rb + 2RbNO3 ==> 6Rb2O + N2
Let us know what you think. Contact Bob Jacobs at
or at the following address: jacobsr@wilton.k12.ct.us