Stoichiometry - AP level
Problems #26-50

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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 times (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 F:

0.128055 g / 19.0 g/mol = 0.00674 mol

4) moles P:

0.2324 g minus 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) get 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 minus the mass of 2 chlorine atoms:
94.88 minus (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 minus 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 minus 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 minus 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

X is lead.


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: More problems as I find them.


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