### Bonus Empirical Formula Problems

These problems represent the next step past doing problems like combustion analysis and the like. Most involve some type of proportion and several use concepts that usually occur later in the first chemistry class. Those of you taking your SECOND chemistry class should be able to do these problems.

Problem #1: Calculate the molar mass of a metal that forms an oxide having the empirical formula M2O3 and contains 68.04% of the metal by mass. Identify the metal.

Problem #2: Hemoglobin is the oxygen carrying compound found in human blood. It is found to contain 0.3335% iron by mass. It is already known that one molecule of hemoglobin contains four atoms of iron. What is the molecular mass of hemoglobin?

Problem #3: For the reaction represented by the equation:

CX4 + 2O2 ---> CO2 + 2X2O

9.0 g of CX4 completely reacts with 1.74 g of oxygen. What is the approximate molar mass of X?

Problem #4: A mixture of NaCl and NaBr weighing 1.234 g is heated with chlorine gas, which converts the mixture completely to NaCl. The total mass of NaCl is now 1.129 g. What are the mass percentages of NaCl and NaBr in the original sample?

Problem #5: A sample of an oxide of vanadium weighing 4.589 g was heated with hydrogen gas to form water and another oxide of vanadium weighing 3.782 g. The second oxide was treated further with hydrogen until only 2.573 g of vanadium metal remained.

(a) What are the simplest formulas of the two oxides?
(b) What is the total mass of water formed in the successive reactions?

Problem #6: The term "alum" refers to a class of compounds of general formula MM*(SO4)2 · 12H2O, where M and M* are different metals. A 20.000 g sample of a certain alum is heated to drive off the water; the anhydrous residue weighs 11.123 g. Treatment of the residue with excess NaOH precipitates all the M* as M*(OH)3, which weighs 4.388 g. Calculate the molar mass of the alum and identity the two metals, M and M*.

Problem #7: 0.158 g of a barium halide is completely precipitated with H2SO4 and 0.124 g of BaSO4 is collected. What is the formula of the barium halide?

Problem #8: A sample of cocaine, C17H21O4N, is diluted with sugar, C12H22O11. When a 1.00 mg sample of this mixture is burned, 2.00 mg CO2 is formed. What is the percent cocaine in the mixture?

Problem #9: To find the formula of a compound composed of iron and carbon monoxide, Fex(CO)y, the compound is burned in pure oxygen, an reaction that proceeds according to the following unbalanced equation.

Fex(CO)y + O2 --> Fe2O3 + CO2

If you burn 1.959 g of Fex(CO)y and obtain 0.799 g of Fe2O3 and 2.200 g of CO2, what is the empirical formula of Fex(CO)y?

Solution

1) Determine mass of Fe in 0.799 g of Fe2O3:

(0.799 g) (111.69 / 159.687) = 0.558845 g

2) Determine mass of C in CO2:

(2.200 g) (12.011 / 44.009) = 0.60043 g

3) Determine moles of each:

Fe ---> 0.558845 g / 55.845 g/mol = 0.0100 mol
C ---> 0.60043 g / 12.011 = 0.04999 = 0.0500 mol

4) Based on the 1:5 molar ratio for Fe:C, we determine the formula for Fex(CO)y to be Fe(CO)5, known as iron pentacarbonyl.

5) We know 0.01 mole of Fe was present as well as 0.05 mole of C. This means 0.05 mole of O was also present. Does this add up to 1.959 grams?

Fe ---> 0.558845 g
C ---> 0.60043 g
O ---> (0.05 mol) (16.00 g/mol) = 0.800 g

0.558845 g + 0.60043 g + 0.800 g = 1.959275 g

Yay!

Problem #10: A substance is 74.34% carbon and 25.66% hydrogen. 250.0 mL of the gaseous substance weighs 0.358 g at STP. Calculate the molecular formula.

Solution:

1) Assume 100 g of compound.

2) Convert grams to moles:

carbon: 74.34 / 12.011 = 6.189
hydrogen: 25.66 / 1.008 = 25.456

3) Convert to smallest whole-number ratio:

carbon: 6.189 / 6.189 = 1
hydrogen: 25.456 / 6.189 = 4.11

4) The proposed empirical formula is CH4; the empirical formula weight is 16.

5) Calculate total moles using PV = nRT:

n = [ (1.000 atm) (0.2500 L) ] / [ (0.08206 L atm / mol K) (273 K) ] = 0.01116 mol

6) Calculate molecular weight of the gas:

0.358 g / 0.01116 mol = 32.1 g/mol

7) Divide molecular weight by empirical formula weight:

32 / 16 = 2

8) The molecular formula is C2H8.

Problem #11: Pure oxygen can be made by heating a compound containing potassium, chlorine and oxygen. What is the empirical formula of this compound, if a 3.22 g sample decomposes to give gaseous oxygen (O2) and 1.96 g KCl?

Solution:

mass of oxygen: 3.22 − 1.96 = 1.26 g
moles of oxygen: 1.26 g / 15.999 g/mol = 0.0788 mol

mass of chlorine: (1.96 g) (35.453 / 74.55) = 0.932 g
moles of chlorine: 0.932 g / 35.453 g/mol = 0.0263 mol

Hint: for potassium, use the same type of calculation as for chlorine, except use the K/KCl gravimetric factor (instead of the Cl/KCl value used just above.

mass of potassium: 1.028 g
moles of potassium: 0.0263 mol

Divide moles by smallest factor (which is 0.0263):

O = 3
Cl = 1
K = 1

The formula is KClO3

Problem #12: A 3.40 g sample of a titanium compound dissolves in water to produce titanium ions and chloride ions. All the chloride ions in the solution are precipitated by the addition of excess silver nitrate solution and after filtration and drying, 9.47 g of the silver chloride was obtained. What is the empirical formula of the compound?

Solution:

1) The gram-atomic weight of chloride ion is 35.453 g/mol. The gram-formula weight of AgCl is 143.323 g/mol

2) The mass of chloride ion in 9.47 g AgCl:

(9.47 g) (35.453 / 143.323) = 2.34254 g

(35.453 / 143.323) is often referred to as a 'gravimetric factor.'

3) The mass of Ti in the original sample:

3.40 g − 2.34254 g = 1.05746 g

4) Divide each mass by respective atomic mass:

Ti ---> 1.05746 / 47.867 = 0.02209 mol
Cl ---> 2.34254 / 35.453 = 0.06607 mol

5) Within experimental error, the Ti to Cl molar ratio is 1:3 The empirical formula is TiCl3

Problem #13: Analysis shows that 2.431 g of Mg reacts with exactly 1.600 g of O2 to form an oxide with the formula MgxOy. What is the formula of the oxide?

Solution:

1) Determine moles of Mg present:

2.431 g / 24.305 g/mol = 0.1000 mol

2) Determine moles of oxygen atoms present:

1.600 g / 31.998 g/mol = 0.05000 mol

(0.0500 mol) (2 atoms/molecule) = 0.1000 mol

3) The molar ratio between Mg and O is 1:1. The formula of the oxide is MgO.

4) Here's another way to think about the 1.600 g of O2:

Allow the 1.600 g of O2 to dissociate into O atoms.

There are now 1.600 g of O atoms.

The gram-atomic weight of O is 16.00 g/mol.

1.600 g / 16.00 g/mol = 0.1000 mol of O atoms

Problem #14: A 13.4 g sample of an unknown liquid is vapourized at 85.0 °C and 100.0 kPa. The vapor has a volume of 4.32 L. The percentage composition of the liquid is found to be 52.1% carbon, 13.2% hydrogen, and 34.7% oxygen. What is the molecular formula?

Solution:

1) Determine the empirical formula of the compound:

C ---> 52.1 g / 12.011 g/mol = 4.3377 mol
H ---> 13.2 g / 1.008 g/mol = 13.095 mol
O ---> 34.7 g / 16.00 g/mol = 2.16875 mol

C ---> 4.3377 / 2.16875 = 2
H ---> 13.095 / 2.16875 = 6.038
O ---> 2.16875 / 2.16875 = 1

Empirical formula ---> C2H6O

2) Determine the moles of the compound:

PV = nRT

(100.0 kPa / 101.325 kPa/atm) (4.32 L) = (n) (0.08206 L atm / mol K) (358 K)

n = 0.14513 mol

3) Determine the molecular weight of the compound:

13.4 g / 0.14513 mol = 92.33 g/mol

4) Determine the molecular formula:

The empirical formula weight of C2H6O is 46.0684

92.33 / 46.0684 = 2

The molecular formula is C4H12O2