### Stoichiometry - AP levelProblems #26-50

Problem #26: A 0.204 gram sample of a metal, M, reacts completely with sulfuric acid according to:

M + H2SO4 ---> MSO4 + H2

A volume of 213 mL of hydrogen is collected over water; the water level in the collecting vessel is the same as the outside level. Atmospheric pressure is 756.0 torr and the temperature is 25.0 °C. Calculate the molar mass of the metal.

The vapor pressure of water at various temperatures can be found in this table.

Solution:

1) Use Dalton's Law of Partial Pressures to determine the pressure of the dry H2:

756.0 − 23.8 = 732.2 torr

2) Converting that to atm gives:

732.2 torr / 760.0 torr/atm = 0.9634 atm

3) Use the ideal gas law:

PV = n RT

(0.9634 atm) (0.213 L) = (n) (0.08206 L atm / mol K) (298 K)

n = 0.0083915 mol H2

4) From the balanced equation, 1 mole of H2 is formed when 1 mol of metal M is reacted. Therefore, the moles of metal that reacted was:

0.0083915 mol M

5) The molar mass of M is:

0.204 g / 0.0083915 mol = 24.3 g/mol

The metal was magnesium.

Problem #27: A common way to obtain a pure metal from its impure metal oxide is to react the oxide with carbon, expressed generically as:

2MO(s) + C(s) ---> 2M(s) + CO2(g)

If 5.00 g of an unknown metal oxide (MO) reacted with excess carbon and formed 738 mL of CO2 at 200.0 °C and 0.978 atm, what is the identity of the metal?

Solution:

1) Using PV = nRT, solve for moles of CO2:

(0.978 atm) (0.738 L) = (n) (0.08206 L atm / mol K) (473 K)

n = 0.0186 mol

2) Determine moles of oxygen atoms in metal oxide:

For every one mole of CO2 produced, two moles of MO were consumed.

1 is to 2 as 0.0186 is to x

x = 0.0372 mol

3) Determine molar mass of MO:

5.00 g / 0.0372 mol = 134.44 mol

4) Determine atomic weight of M:

x + 16.00 = 134.44

x = 118.44 g/mol

M is tin.

Problem #28: A compound of P and F was analyzed as follows: heating 0.2324 g of the compound in a 378 cm3 flask turned all of it to gas, which had a pressure of 97.3 mmHg at 77 °C. Then, the gas was mixed with calcium chloride solution which turned all of the F to 0.2631 g of CaF2. Determine the molecular formula of the compound.

Solution:

1) Molecular weight of the gas:

(97.3 / 760) (0.378) = (n) (0.08206) (350)

n = 0.001684967 mol

0.2324 g / 0.001684967 mol = 138 g/mol

We will use this to move from the empirical formula to the molecular formula.

2) Mass of F:

(0.2631 g) (38.0 g / 78.074 g) = 0.128055 g

(38.0 g / 78.074 g) is called a gravimetric factor. It is the decimal percent of F in CaF2.

3) Moles of F:

0.128055 g / 19.0 g/mol = 0.00674 mol

4) Moles of P:

0.2324 g − 0.128055 g = 0.104345 g

0.104345 g / 31.0 g/mol = 0.003366 mol

5) Empirical formula:

P ---> 0.003366 / 0.003366 = 1
F ---> 0.00674 / 0.003366 = 2

PF2

6) Molecular formula:

PF2 weighs 31 + 38 = 69

138 / 69 = 2

PF2 times 2 = P2F4

Another P2F4 problem.

Problem #29: A metal chloride reacts with silver nitrate solution to give a precipitate of silver chloride according to following equation:

MCl2 + 2AgNO3 ---> M(NO3)2 + 2AgCl

When a solution containing 0.4750 g of metal chloride is made to react with silver nitrate, 1.435 grams of silver chloride are formed. Identify the metal.

Solution:

1) Determine moles of AgCl produced:

1.435 g AgCl / 143.32 g/mol = 1.001256 x 10-2 mol AgCl

2) Determine moles of MCl2 consumed:

1.001256 x 10-2 mol AgCl x (1 mol MCl2 / 2 mol AgCl) = 5.00628 x 10-3 mol MCl2

3) Determine molar mass of MCl2:

0.4750 g / 5.00628 x 10-3 mol = 94.88 g/mol

4) Determine the atomic weight and identity of M:

The atomic of M will be:

94.88 − (2 x 35.453) = 23.97 g/mol

M is magnesium

Problem #30: An unidentified metal M reacts with an unidentified halogen X to form a compound MX2. When heated the compound decomposes by the reaction:

2MX2 (s) ---> 2MX (s) + X2 (g)

When 1.12 g of MX2 is heated, 0.720 g of MX is obtained along with 56.0 mL of X2 gas (at STP).

a) What is the atomic mass and the identity of the halogen X?
b) What is the atomic mass and identity of the metal M?

Solution:

1) Calculate moles of X2 collected:

PV = nRT

(1.00 atm) (0.0560 L) = (n) (0.08206 L atm / mol K) (273 K)

n = 0.00250 mol

2) Calculate mass of X2 collected:

1.12 g − 0.720 g = 0.40 g

3) Calculate molecular weight of X2 and determine identity of X

0.40 g/0.00250 moles = 160 g/mol

half of 160 is 80, the atomic weight of X

bromine

4) Determine moles of MX:

MX : X2 molar ratio is 2 : 1

therefore, 0.0050 moles of MX

5) Calculate molar mass of MX:

0.720 g/0.0050 moles = 144 g/mole

6) Calculate atomic weight of M:

144 g/mol − 80 g/mol = 64 g/mole

copper

Problem #31: A metal sulfate has the formula M2SO4. 10.99 g of the compound was dissolved in water to make 500.0 cm3 of solution. A 25.0 cm3 sample was removed and reacted with an excess of BaCl2(aq) to produce a precipitate of BaSO4, which when dried had a mass of 1.167 g.

a) Determine the number of moles of BaSO4 precipitated.
b) Determine the concentration of M2SO4
c) Identify M

Solution:

1) Moles of BaSO4 precipitated from 25 cm3:

1.167 g / 233.4 g/mol = 0.005 mol

2) Moles of BaSO4 that would have precipitated from 500 cm3:

0.005 mol is to 25 as x is to 500

x = 0.100 mol of BaSO4

3) The molarity of the M2SO4 is this:

0.100 mol / 0.500 L = 0.200 M

The 0.100 mole comes from the fact that there is a 1:1 molar ratio between M2SO4 and BaSO4

M2SO4 + CaCl2 ---> BaSO4 + 2MCl

4) What is M?

0.1 mol of sulfate weighs 96.061 x 0.1 = 9.6061 g

10.99 − 9.6061 = 1.3839 g of M+ in solution

There were 0.2 mol of M+ in the 500 cm3 [from this: M2SO4(s) ---> 2M+(aq) + SO42-(aq)]

1.3839 g / 0.2 mol = 6.9 g/mol

Li2SO4

Problem #32: An element, X, forms two compounds with bromine: XBr2 and XBr4. When 10.00 grams of the XBr2 is reacted with excess bromine, 14.35 g of XBr4 is formed. Identify X.

Solution:

14.35 − 10.00 = 4.35 g of Br2 reacted

4.35 g / 159.808 g/mol = 0.02722 mol of Br2

XBr2 + Br2 ---> XBr4

XBr2 and Br2 react in a 1:1 molar ratio

Therefore, 10.00 g represents 0.02722 mol of XBr2

10.00 g / 0.02722 mol = 367 g/mol

In 367.377 g of XBr2, there is 159.808 g of bromine

367.377 − 159.808 = 207.569 <--- this is the atomic weight of X

Problem #33: Exactly 4.32 g of oxygen gas was required to completely combust a 2.16 g sample of a mixture of methanol and ethanol:

(1) How many moles of ethanol are contained within the sample?
(2) What is the percentage by weight of methanol in the sample?

Solution:

1) Balanced chemical equations:

CH3OH + 1.5O2 ---> CO2 + 2H2O
C2H5OH + 3O2 ---> 2CO2 + 3H2O

2) Let x = mass of methanol; let y = mass of ethanol

x + y = 2.16

[(1.5) (x/32)] + [(3) (y/46)] = 4.32/32

3) Using x = 2.16 − y:

[(1.5) ((2.16-y)/32)] + [(3) (y/46)] = 0.135

[(3.24 − 1.5y) / 32] + [3y / 46] = 0.135

4) Multiply each side by 32, then by 46:

(46) (3.24 − 1.5y) + (32) (3y) = 198.72

149.04 − 69y + 96y = 198.72

27y = 49.68

y = 1.84 g

x = 0.32 g

5) Answers to (1) and (2):

moles ethanol ---> 1.84 g / 46 g/mol = 0.040 mol

percent methanol ---> (0.32 g / 2.16 g) * 100 = 14.8%

Problem #34: A 3.41 g sample of a metallic element, M, reacts completely with 0.0158 mol of a gas, X2, to form 4.53 g MX. What are the identities of M and X?

Solution:

2M + X2 ---> 2MX

We know that M and X2 react in a 2:1 molar ratio.

Therefore, 0.0158 mol of X2 reacts with twice that many moles of M, 0.0316 mol

We can now determine the atomic weight of M:

3.41 g / 0.0316 mol = 107.9 g/mol

From the periodic table, we see that M is silver.

4.53 − 3.41 = 1.12 g <--- this is the mass of X in MX

We know that there is a 1:1 molar ratio between M and MX, therefore 0.0316 of MX was produced. Since we know the formula is MX, we know that 0.0316 mol of X is involved.

1.12 g / 0.0316 mol = 35.44 g/mol

This atomic weight is within experimental error for X to be chlorine.

Problem #35: When 2.3 moles of X reacts with 1.6 moles of Y, 71 grams of Z are produced. What is the molar mass of Z?

3X + 4Y ---> 5Z

This reaction has a 50% yield.

Solution:

1) Determine the limiting reagent:

X: 2.3 / 3 = 0.77
Y: 1.6 / 4 = 0.4

Y is the limiting reagent.

2) How much Z is produced at 100% yield:

4 is to 5 as 1.6 is to x

x = 2.0 mol

3) The reaction has only 50% yield, so 1.0 mole of Z was produced. The molar mass of Z is:

71 g / 1.0 mol = 71 g/mol

Problem #36: 20.0 mL of solution containing NaCl and KCl gave, on evaporation to dryness, 0.180 g of the mixed chlorides. 20.0 mL of the same solution gave 0.370 g of AgCl on treatment with a slight excess of the AgNO3 solution. Calculate, for the original solution, the mass per liter of both chlorides.

Solution:

1) Let x = mass NaCl and let y = mass KCl. from that, we get our first equation:

x + y = 0.180

2) Using AgCl, determine the moles of chloride ion in the solution:

moles AgCl ---> 0.370 g / 143.321 g/mol = 0.00258 mol

0.00258 mol gives the moles of chloride based on this:

Ag+(aq) + Cl¯(aq) ---> AgCl(s)
For every 1 mole of AgCl that precipitated, there was 1 mole of chloride in solution.

3) We can now write our second equation:

x / 58.4428 g/mol + y / 74.553 = 0.00258

x / 58.4428 ---> moles of NaCl
y / 74.553 ---> moles of KCl

Based on the 1:1 molar ratios between NaCl and Cl¯ (as well as KCl and Cl¯), we know that the two above divisions give the moles of chloride contributed from the NaCl and the KCl.

4) We will now substitute the first equation into the second and solve:

x = 0.180 − y

[(0.180 − y) / 58.4428] + [y / 74.553] = 0.00258

13.4 − 74.553y + 58.4428y = 11.2

16.1y = 2.2

y = 0.137 g <--- mass KCl in 20.0 mL
x = 0.180 − 0.137 = 0.0430 g <--- mass NaCl in 20.0 mL

mass KCl in 1 L ---> 0.137 g / 0.020 L = 6.85 g
mass NaCl in 1 L ---> 0.043 / 0.020 L = 2.15 g

Problem #37: You are given a mixture of three hydrated salts: Na2CO3 · 10H2O, MgSO4 · 7H2O, and CuSO4 · 5H2O. The total mass of the mixture is 12.123 grams. When the mixture is heated gently, the following two reactions occur:

Na2CO3 · 10H2O(s) ---> Na2CO3 · 7H2O(s) + 3H2O(g)
MgSO4 · 7H2O(s) ---> MgSO4 · H2O(s) + 6H2O(g)

After these reactions are complete, the mass of the mixture has decreased to 9.049 grams. This mixture is then heated more strongly, and the following additional reactions occur:

Na2CO3 · 7H2O(s) ---> Na2CO3(s) + 7H2O(g)
MgSO4 · H2O(s) ---> MgSO4(s) + H2O(g)
CuSO4 · 5H2O(s) ---> CuSO4(s) + 5H2O(g)

After this final heating, the mass of the mixture has decreased to 6.412 grams. From this information, calculate the masses of each of the three compounds in the original mixture.

Solution:

Comment: three simultaneous equations in three unknowns are required.

1) The first equation:

X + Y + Z = 12.123 g

where:

X = the mass of Na2CO3 · 10H2O
Y = the mass of MgSO4 · 7H2O
Z = the mass of CuSO4 · 5H2O

2) The second equation is developed based on the information gained from the first heating. Here is the second equation:

(232.0916 / 286.136) (X) + (138.3808 / 246.4696) (Y) + Z = 9.049 g

(232.0916 / 286.136) ---> this gravimetric factor is the decimal percent decrease in the original mass (the X) of the sodium carbonate decahydrate as it is changed to the heptahydrate. 232 is the molar mass of the heptahydrate and 286 is the molar mass of the decahydrate.

(138.3808 / 246.4696) ---> this gravimetric factor is the decimal percent decrease in the original mass (the Y) of magnesium heptahydrate as it is changed to the monohydrate. 246 is the molar mass of the heptahydrate and 138 is the molar mass of the monohydrate

There is no gravimetric factor for copper(II) sulfate pentahydrate since it did not lose any mass. It started with Z grams present and ended with Z grams present.

By the way, after the heating, the sample is now composed of Na2CO3 · 7H2O, MgSO4 · H2O, and CuSO4 · 5H2O.

3) The third equation is developed based on the information gained from the second heating. Here is the third equation:

(105.988 / 286.136) (X) + (120.366 / 246.4696) (Y) + (159.607 / 249.681) (Z) = 6.412 g

Notice that, in each gravimetric factor, the denominator is the molar mass of the original hydrate. This is because the 6.412 value references the entire loss of mass from the starting value of 12.123 g.

4) I'm going to rewrite the three equations, but I will use decimals rather than fractions:

X + Y + Z = 12.123

0.8111234X + 0.5614518Y + Z = 9.049

0.3704113X + 0.4883604Y + 0.6392437Z = 6.412

5) To calculate the answers, I used a 3 Equation System Solver. When I did that, I got the following answers (which I rounded off):

1.373 grams of Na2CO3 · 10H2O
6.418 grams of MgSO4 · 7H2O
4.332 grams of CuSO4 · 5H2O

6) I will start a step-by-step solution:

Rewrite the first equation:
Z = 12.123 − (X + Y)

Substitute into the second and third equations:

0.8111234X + 0.5614518Y + [12.123 − (X + Y)] = 9.049

0.3704113X + 0.4883604Y + 0.6392437[12.123 − (X + Y)] = 6.412

We now have two simultaneous equations in 2 unknowns.

And that is where I will leave it.

7) I must admit that I did not solve this problem on my own. When I found the problem in my notes, I did an Internet search which yielded some discussion about this problem. Look for links to 'www.chemicalforums.com'

Problem #38: A mixture of CuSO4 · 5H2O and MgSO4 · 7H2O is heated until all the water is lost. If 5.020 g of the mixture gives 2.988 g of the anhydrous salts, what is the percent by mass of CuSO4 · 5H2O in the mixture?

Solution:

1) For every mole of CuSO4 there are 5 moles of H2O, and for every mole of MgSO4 there are 7 moles of H2O. Therefore:

(5 * moles of CuSO4) + (7 * moles of MgSO4) = moles of H2O

2) Determine moles of water lost:

5.020 − 2.988 = 2.032 g

2.032 g / 18.015 g/mol = 0.112795 mol

3) Let X be the grams of anhydrous CuSO4. Therefore:

2.988 − X ---> the grams of anhydrous MgSO4

3) We can now substitute into the equation in step 1:

5 * (X / 159.6096) + 7 * [(2.988 − X) / 120.3686] = 0.112795

X / 159.6096 ---> moles of anhydrous CuSO4
(2.988 − X) / 120.3686 ---> moles of anhydrous MgSO4

4) Solve it:

(159.6096) (120.3686) * 5 * (X / 159.6096) + (159.6096) (120.3686) * 7 * [(2.988 − X) / 120.3686] = (0.112795) (159.6096) (120.3686)

(120.3686) * 5 * X + (159.6096) * 7 * (2.988 − X) = 2167.016

601.843X + 3338.3944 − 1117.2672X = 2167.016

515.4242X = 1171.3784

x = 2.27265 g of anhydrous CuSO4

5) Determine moles of anhydrous CuSO4:

2.27265 g / 159.6096 g/mol = 0.0142388 mol

6) Determine mass of water associated with 0.0142388 mol of anhydrous CuSO4:

(0.0142388 mol) (5) = 0.071194 mol of H2O

(0.071194 mol) (18.015 g/mol) = 1.28256 g

7) Determine mass of CuSO4 · 5H2O in the original sample and its percentage:

2.27265 g + 1.28256 g = 3.55521 g

(3.55521 / 5.020) * 100 = 70.82%

Problem #39: One mole of hydrocarbon Z is subjected to combustion. The product obtained is condensed and the resulting gaseous product occupied a volume of 89.6 L at STP. Oxygen required for this combustion was 145.6 L at STP. What is the molecular formula of Z?

Solution:

1) Let the formula for Z be this:

CxHy

2) Determine moles of CO2 and O2:

89.6 L / 22.4 L/mol = 4.00 mol CO2

145.6 L / 22.4 L/mol = 6.50 mol O2

3) The reaction thus far:

CxHy + 6.5O2 ---> 4CO2 + ___H2O

4) Determine the coefficient for H2O:

6.50 − 4.00 = 2.50 mol of O2 in the H2O

giving 5.00 mol H2O

5) Therefore:

CxHy + 6.5O2 ---> 4CO2 + 5H2O

x must be 4 and y must be 10

C4H10