### 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: 2.50 g calcium metal is placed in a solution of chromium acetate. After 20 minutes, 1.30 g of un-reacted calcium metal remained and was removed from the solution. If 1.04 g of chromium metal precipitated during the reaction, what is the empirical formula of the chromium acetate in the solution?

Solution:

grams calcium ---> 2.50 − 1.30 = 1.20 g
moles calcium ---> 1.20 g / 40.08 g/mol = 0.02994 mol (real close to 0.03)

moles Cr ---> 1.04 g / 51.9961 g/mol = 0.02000 mol

Ca + Cr?+ ---> Ca2+ + Cr

One Ca loses two electrons, so 0.03 mole of Ca oxidized liberates 0.06 mol of electrons.

That 0.06 mole of electrons reduces 0.02 mole of Cr.

Cr must be +3

Cr(C2H3O2)3

Problem #15: The observed heat capacity per gram of a compound containing rubidium and oxygen is 0.64 J K¯11. Use the Law of Dulong and Petit to determine the empirical formula of the compound.

Solution:

1) Dulong-Petit:

the heat capacity of a solid is 3R per mol of atoms.

Note: this is not moles of the compound, but total moles of atoms.

2) Let the formula be RbxOy. When we have one gram of it, the moles of atoms is approx:

(1 gram / (85x + 16y grams per mole)) * ( x + y )

This factor:

1 gram / (85x + 16y grams per mole)

computes the amount of moles of RbxOy in one gram. When that value is multiplied by the number of atoms in one formula unit (the x +y), we have the total number of moles of atoms.

3) The heat capacity of one gram of RbxOy is equal to:

3R * (1 gram / (85x + 16y grams per mole)) * ( x + y ) = 0.64 J K¯11

where R = 8.31447 J mol¯11

4) By trial and error, the x and y values can be determined. However, x and y must both be small, whole numbers.

when x is set to 1 and y is set to 2, the left-hand side of the above equation is approx. 0.6396

5) The compound is RbO2, rubidium superoxide. An unusual compound, to say the least. An analogous compound would potassium superoxide, KO2.

Problem #16: Calculate the empirical formula of hydrated ferrous ammonium sulfate, Fea(NH4)b(SO4)c(H2O)d, from the following data.

(a) 0.7840 g of the salt gives 0.1600 g Fe2O3(s) when heated strongly in air to constant mass.

(b) 0.7840 g of the salt dissolved in water gives 0.9336 g BaSO4(s) when excess BaCl2(aq) is added.

(c) When 0.3920 g of the salt is dissolved in water and boiled with excess NaOH(aq), NH3(g) is liberated. When this gas is absorbed in 50.0 mL of 0.10 M HCl(aq), the excess acid remaining after reaction with the NH3(g) requires 30.0 mL of 0.10 M NaOH for neutralization.

Solution:

1) Each component of the hydrated ferrous ammonium sulfate will have its percent composition determined:

(a) Iron:
(0.1600 g) (111.69 / 159.687) = 0.1119089 g (mass of Fe in sample)

(0.1119089 g / 0.7840) (100) = 14.274% (percent composition of Fe in hydrated ferrous ammonium sulfate)

(b) Sulfate (treated as an SO4 group, not individual sulfur and oxygen):

(0.9336 g) (96.061 / 233.391) = 0.384259 g

(0.384259 g / 0.7840 g) (100) = 49.013% (percent composition of sulfate in hydrated ferrous ammonium sulfate)

(c) Ammonium:

(0.10 mol/L) (0.0300 L) = 0.0030 mol (NaOH used in back titration)

By 1:1 molar ratio between NaOH and HCl, we have 0.0030 mol of excess HCl after NH3 reacted

(0.10 mol/L) (0.0500 L) = 0.0050 mol (total HCl present before reaction with NH3)

0.0050 mol − 0.0030 mol = 0.0020 mol (HCl that reacted with the ammonia)

By 1:1 molar ratio between HCl and NH3, we have 0.0020 mol of NH3 that was liberated from the hydrated ferrous ammonium sulfate

There is a 1:1 molar ratio between ammonia liberated and ammonium in the sample.

(0.0020 mol) (18.0386 g/mol) = 0.0360772 g (mass of ammonium in sample)

(0.0360772 g / 0.3920 g) (100) = 9.20337% (percent composition of ammonium in hydrated ferrous ammonium sulfate

(d) Water:

100 − (14.274 + 49.013 + 9.20337) = 27.50963%

2) Assume 100 g of the compound is present. Convert masses to moles:

Fe ---> 14.274 g / 55.845 g/mol = 0.2556 mol
SO42¯ ---> 49.013 g / 96.061 g/mol = 0.51 mol
NH4+ ---> 9.20337 g / 18.0386 g/mol = 0.51 mol
H2O ---> 27.50963 g / 18.0152 g/mol = 1.527 mol

3) Divide by smallest value:

Fe ---> 0.2556 mol / 0.2556 mol = 1
SO42¯ ---> 0.51 mol / 0.2556 mol = 2
NH4+ ---> 0.51 mol / 0.2556 mol = 2
H2O ---> 1.527 mol / 0.2556 mol = 6

4) Write the empirical formula: Fe(NH4)2(SO4)2(H2O)6