### Combustion Analysis: Problems 1 - 10

Problem #1: 0.487 grams of quinine (molar mass = 324 g/mol) is combusted and found to produce 1.321 g CO2, 0.325 g H2O and 0.0421 g nitrogen. Determine the empirical and molecular formulas.

Problem #2: 95.6 mg of menthol (molar mass = 156 g/mol) are burned in oxygen gas to give 269 mg CO2 and 110 mg H2O. What is menthol's empirical formula?

Problem #3: 0.1005 g of menthol is combusted, producing 0.2829 g of CO2 and 0.1159 g of H2O. What is menthol's empirical formula? (Yes, the answer will be the same as #12.)

Problem #4: The combustion of 40.10 g of a compound which contains only C, H, Cl and O yields 58.57 g of CO2 and 14.98 g of H2O. Another sample of the compound with a mass of 75.00 g is found to contain 22.06 g of Cl. What is the empirical formula of the compound?

Solution #1:

1) Get grams of each element:

Carbon: (58.57 g) (12.011 / 44.009) = 15.985 g of C in 40.10 g sample
Hydrogen: (14.98 g) (2.016 / 18.015) = 1.6764 g of H in 40.10 g sample
Oxygen: we leave this to later (see below for an interesting solution path that involves determining the mass of oxygen.)
Chlorine: problem gives 22.06 g in 75.00 g sample

2) Let us determine the percent composition:

Carbon: 15.985 g / 40.10 g = 39.86%
Hydrogen: 1.6764 g / 40.10 = 4.18%
Chlorine: 22.06 / 75.00 = 29.41%
Oxygen: 100% − (39.86 + 4.18 +2 9.41) = 26.55%

3) Let us assume 100 g of the compound. In which case, the percentages above become grams. Now, let us determine the moles of each (I'll skip typing the calcs):

C: 3.32 mol
H: 4.147 mol
O: 1.66 mol
Cl: 0.83 mol

4) Divide each by 0.83

C: 4
H: 5
O: 2
Cl: 1

The empirical formula is C4H5ClO2

Solution #2:

1) Determine mass of all four compounds in the 40.10 g sample:

Carbon: 15.985 g
Hydrogen: 1.676 g

Here comes the interesting way that is different from solution #1:

Chlorine: (40.10 g) (22.06 g Cl / 75.00 g sample) = 11.795 g
Oxygen: 40.10 g − 15.985 g − 1.676 g − 11.795 g = 10.644 g

Pretty slick, heh? Notice how the oxygen is determined by subtraction after everything else is calculated. This is a common pattern in combustion analysis.

2) Determine the moles of each (I'll skip typing the calcs):

C = 1.331 mol
H = 1.663 mol
Cl = 0.3327 mol
O = 0.6653 mol

Note that there is no need to assume 100 g of the compound and work from the percent composition.

3) Divide all by the smallest to simplify:

C = 4
H = 5
Cl = 1
O = 2

The empirical formula is C4H5ClO2

Problem #5: The combustion of 1.38 grams of a compound which contains C, H, O and N yields 1.72 grams of CO2 and 1.18 grams of H2O. Another sample of the compound with a mass of 22.34 grams is found to contain 6.75 grams of O. What is the empirical formula of the compound?

1) Calculate grams of C and H:

carbon: (1.72 g) (12.011 / 44.009) = 0.4694 g of C
hydrogen: (1.18 g) (2.016 / 18.015) = 0.13205 g of H
oxygen: see next step
nitrogen: see next step

2) Calculate mass percent of each element:

carbon: 0.4694 g / 1.38 g = 34.01%
hydrogen: 0.13205 g / 1.38 g = 9.57%
oxygen: 6.75 g / 22.34 g = 30.215%
nitrogen: 100 − (34.01 + 9.57 + 30.215) = 26.205%

3) Assume 100 g of compound present. Therefore:

carbon: 34.01 g
hydrogen: 9.57 g
oxygen: 30.215 g
nitrogen: 26.205 g

4) Calculate moles:

carbon: 34.01 g / 12.011 g/mol = 2.832
hydrogen: 9.57 g / 1.008 g/mol = 9.494
oxygen: 30.215 g / 16.00 g/mol = 1.888
nitrogen: 20.205 g / 14.007 g/mol = 1.871

5) Look for smallest, whole-number ratio:

carbon: 2.832 / 1.871 = 1.5 (x 2 = 3)
hydrogen: 9.494 / 1.871 = 5 (x 2 = 10)
oxygen: 1.888 / 1.871 = 1 (x 2 = 1)
nitrogen: 1.871 / 1.871 = 1 (x 2 = 2)

6) Empirical formula:

C3H10N2O2

Problem #6: The combustion of 3.42 g of a compound is known to contain only nitrogen and hydrogen gave 9.82 g of NO2 and 3.85 g of water. Determine the empirical formula of this compound.

Solution:

1) Calculate moles of N and moles of H in the combustion products:

Moles N
9.82 g NO2 / 46.0 g/mol = 0.213 mol NO2
(0.213 mol NO2) (1 mol N / 1 mol NO2) = 0.213 mol N

Moles H
3.85 g H2O / 18.0 g/mol = 0.213 mol H2O
(0.213 mol H2O) (2 mol H / 1 mol H2O ) = 0.428 mol H

2) Calculate the ratio of moles by dividing both by the smaller:

N ---> 0.213 / 0.213 = 1
H ---> 0.428 / 0.213 = 2

The empirical formula is NH2

Problem #7: A compound with a known molecular weight (146.99 g/mol) that contains only C, H, and Cl was studied by combustion analysis. When a 0.367 g sample was combusted, 0.659 g of CO2 and 0.0892 g of H2O formed. What are the empirical and molecular formulas?

Solution:

1) Carbon:

0.659 g of CO2 has 0.659 / 44 = 0.0150 moles of CO2

there is 1 mole of C in CO2 and all the C from the compound becomes CO2, so moles of C in the compound = 0.0150 moles

mass of C = 0.0150 x 12 = 0.1797 g

2) Hydrogen:

0.089 g of H2O has 0.0892 / 18 = 0.0050 moles of H2O

there are 2 moles of H in H2O, so moles of H in the compound = 0.0099 moles

mass of H = 0.0099 x 1.0079 = 0.0100 g

3) Chlorine:

mass of H + C = 0.1897 g
mass of sample = 0.3670 g
mass of Cl by difference = 0.1773 g
moles of Cl = 0.0050 moles

4) Smallest whole-number ratio:

molar ratio of C : H : Cl = 0.0150 : 0.0099 : 0.0050
divide the ratio by the smallest number
molar ratio of C : H : Cl = 3.00 : 1.98 : 1

5) Formulas:

empirical formula is C3H2Cl

this has an "empirical formula weight" of (36+2+35.5) = 73.5 g

which is 1/2 the molecular mass

so the molecular formula is twice the empirical formula

C6H4Cl2

Problem #8: A 2.52 g sample of a compound containing carbon, hydrogen, nitrogen, oxygen, and sulfur was burned in excess oxygen gas to yield 4.36 grams of CO2 and 0.892 grams of H2O as the only carbon and hydrogen products respectively. Another sample of the same compound of mass 4.14 g yielded 2.60 g of SO3 as the only sulfur containing product. A third sample of mass 5.66 g was burned under different conditions to yield 2.80 g of HNO3 as the only nitrogen containing product. Determine the empirical formula of the compound.

Solution #1:

1) Carbon:

(4.36 g) (12.011 g / 44.0 g) = 1.1902 g of C
1.1902 g / 2.52 g = 47.23%

2) Hydrogen:

(0.892 g) (2.016 g / 18.015 g) = 0.09982 g of H
0.09982 g / 2.52 g = 3.96%

3) Sulfur:

2.60 g x (32.065 g / 80.062 g) = 1.0413 g of S
1.0413 g / 4.14 g = 25.15%

4) Nitrogen:

2.80 x (14.007 / 63.012) = 0.6224 g of N
0.6224 g / 5.66 g = 11.00%

5) Oxygen:

100% − (47.23% + 3.96% + 25.15% + 11.00%) = 12.66%

6) Assume 100 g of compound present:

C = 47.23 g; H = 3.96 g; S = 25.15 g; N = 11.00 g; O = 12.66 g

7) Convert to moles:

C = 3.93; H = 3.93; S = 0.7843; N = 0.7853; O = 0.79125

8) Convert to lowest whole-number ratio by dividing each mole amount by 0.7843:

C = 5; H = 5; S = 1; N = 1; O = 1

Empirical formula is C5H5NOS

Repeat Problem #8: A 2.52 g sample of a compound containing carbon, hydrogen, nitrogen, oxygen, and sulfur was burned in excess oxygen gas to yield 4.36 grams of CO2 and 0.892 grams of H2O as the only carbon and hydrogen products respectively. Another sample of the same compound of mass 4.14 g yielded 2.60 g of SO3 as the only sulfur containing product. A third sample of mass 5.66 g was burned under different conditions to yield 2.80 g of HNO3 as the only nitrogen containing product. Determine the empirical formula of the compound.

Solution #2:

1) Carbon:

# moles = mass / molar mass
molar mass of CO2 = 44.0 g/ mole
4.36 g of CO2 has 4.36 /44.0 = 0.09909 moles of CO2
there is 1 mole of C in CO2 and all the C from the compound becomes CO2
moles of C in the compound = 0.09909 moles
mass of C ---> 0.09909 x 12 = 1.1891 g

2) Hydrogen:

molar mass of H2O = 18 g/ mole
0.892 g of H2O has 0.892 / 18 = 0.04956 moles of H2O
there are 2 moles of H in H2O so moles of H in the compound = 0.09911 moles
mass of H ---> 0.09911 x 1.0079 = 0.100 g

3) Sulfur:

moles of SO3 ---> 2.60 / 80 = 0.0325
moles of S in 4.14 g of compound = 0.0325 mol
moles in 2.52 g of compound ---> (0.0325) (2.52 / 4.14) = 0.01978 moles
mass of S ---> 0.01978 x 32 = 0.6330 g

All the sulfur in the SO3 came from the 4.14 g sample.
Notice the scaling from 4.14 g of compound to 2.52 g.

4) Nitrogen:

moles of HNO3 ---> 2.80 / 63 = 0.04444 mol
moles of N in 5.66 g of sample = 0.04444
moles in 2.52 g of compound ---> (0.04444) (2.52 / 5.66) = 0.01979 moles
mass of N ---> 0.01979 x 14 = 0.2770 g

All the nitrogen in the HNO3 came from the 5.66 g sample.
Notice the scaling from 5.66 g of compound to 2.52 g.

5) Oxygen:

mass C + H + S + N = 1.1891 + 0.100 + 0.6330 + 0.2770 = 2.1991
mass of O by difference = 2.52 − 2.1991 = 0.3209 g
moles of O in 2.52 g = 0.3209 / 16 = 0.0201 moles

6) We now have all five mole amounts, so do the empirical formula:

molar ratio of C : H : S : N : O = 0.09909 : 0.09911 : 0.01978 : 0.01979 : 0.0201
divide by the smallest number to get whole-number ratio
C : H : S : N : O = 5 : 5 : 1 : 1 : 1

empirical formula is C5H5NOS

Problem #9: Burning 11.2 mL (measured at STP) of a gas known to contain only carbon and hydrogen, we obtained 44.0 mg CO2 and 0.0270 g H2O. Find the molecular formula of the gas.

Solution:

1) Determine mass of carbon and hydrogen:

C: (0.0440 g) (12.011 / 44.01) = 0.0120 g
H: (0.0270 g) (2.016 / 18.015) = 0.0030 g

2) Determine moles of carbon and hydrogen:

C: 0.0120 g / 12.0 g/mol = 0.00100 mol
H: 0.0030 g / 1.008 g/mol = 0.00300 mol

3) Determine lowest whole-number ratio:

C: 0.00100 mol / 0.00100 mol = 1
H: 0.00300 mol / 0.00100 mol = 3

empirical formula = CH3

4) Determine how many moles are in our 11.2 mL of gas:

PV = nRT

(1.00 atm) (0.0112 L) = (n) (0.08206) (273 K)

n = 0.00050 mol

5) The gas sample weighed this:

0.012 g + 0.003 g = 0.015 g

6) Get molecular weight of gas:

0.015 g / 0.00050 mol = 30 g/mol

7) The "empirical formula weight" of CH3 = 15

30 / 15 = 2

The molecular formula is C2H6

Problem #10: The osmotic pressure of a solution containing 2.04 g of an unknown molecular compound dissolved in 175.0 mL of solution at 25.0 °C is 2.13 atm. The combustion of 22.08 g of the unknown compound produced 36.26 g CO2 and 14.85 g H2O

Solution:

1) The osmotic pressure will allow us to calculate the molar mass of the substance:

π = iMRT

2.13 atm = (1) (x / 0.175 L) (0.08206 L atm / mol K) (298 K)

x = 0.015243 mole

2.04 g / 0.015243 mole = 133.83 g/mol

2) Let us calculate the amount of carbon and hydrogen. Then, by subtraction, we will check for oxygen:

carbon ---> 36.26 g x (12.011 / 44.01 ) = 9.8959 g
hydrogen ---> 14.85 g x (2.016 / 18.015 ) = 1.6618 g
oxygen ---> 22.08 minus (9.8959 + 1.6618) = 10.5223 g

3) Calculate moles of each element:

carbon ---> 9.8959 g / 12.011 g/mol = 0.8239 mol
hydrogen ---> 1.6618 g / 1.008 g/mol = 1.6486 mol
oxygen ---> 10.5223 g / 15.999 g/mol = 0.657685 mol

4) Determine a whole number ratio:

carbon ---> 0.8239 / 0.657685 = 1.25
hydrogen ---> 1.6486 / 0.657685 = 2.5
oxygen ---> 0.657685 / 0.657685 = 1

Look at it like this:

carbon ---> 1.25 = 5/4 (times 4 = 5)
hydrogen ---> 2.5 = 10/4 (times 4 = 10)
oxygen ---> 1 = 4/4 (times 4 = 4)

5) The empirical formula is:

C5H10O4

6) The "empirical formula weight" is 134. We calculated a molecular weight of 133.83. The molecular formula is:

C5H10O4

Problem #11: 1.5 L of a gaseous compound consisting of carbon and hydrogen is combusted with oxygen. When the resulting gaseous products are measured at the initial temperature and pressure, it was found that 3.0 L of CO2 and 1.5 L of H2O were formed. What is the formula of the compound?

Solution:

Because everything was measured at equal temperatures and pressures, Avogadro's Hypothesis can be used to solve this problem. The hypothesis is: "Equal volumes at equal temperatures and pressures contain equal number of molecules."

Just to make it a bit simpler to explain, let us assume a value of T and P such that the 3.0 L of each gas contains 3.0 moles. That means 1.5 moles of the compound was initially present.

3.0 moles of CO2 contains 3.0 moles of C and 1.5 moles of H2O contains 3.0 moles of H.

3.0 moles of C and 3.0 moles of H are contained in 1.5 moles of the compound. Therefore, 1.0 mol of the compound contains 2.0 mol of C and 2.0 mol of H.

The formula of the compound is C2H2.

Bonus Problem #1: A 6.20-g sample of an unknown compound containing only C, H, and O combusts in an oxygen rich environment. When the products have cooled to 20.0 °C at 1 bar, there are 8.09 L of CO2 and 3.99 mL of H2O. The density of water at 20.0 °C is 0.998 g/mL.

a) What is the empirical formula of the unknown compound?
b) If the molar mass is 168.2 g/mol, what is the molecular formula of the compound?

Solution:

1) Mass of H2O:

(3.99 mL) (0.998 g/mL) = 3.982 g

2) Mass of CO2:

PV = nRT

(1 bar) (8.09 L) = (n) (0.0831447 L bar / mol K) (293 K)

n = 0.33208276 mol

(0.33208276 mol) (44.009 g/mol) = 14.61463 g

3) We need grams of C and H before getting grams of O:

(14.61463 g) (12.011 g / 44.009 g) = 3.988646 g

(3.982 g) (2.016 g / 18.015 g) = 0.445613 g

4) Mass of O:

6.20 − (3.988646 + 0.445613) = 1.765741 g

5) Determine moles of all three components:

3.988646 g / 12.011 g/mol = 0.332083 mol
0.445613 g / 1.008 g/mol = 0.4420764 mol
1.765741 g / 16.00 g/mol = 0.110359 mol

6) Divide through by smallest:

0.332083 mol / 0.110359 mol = 3
0.4420764 mol / 0.110359 mol = 4
0.110359 mol / 0.110359 mol = 1

Empirical formula is C3H4O

7) The weight of C3H4O is 56.0636

168.2 / 56.0636 = 3

Molecular formula is C9H12O3

Bonus Problem #2: Given the following reaction:

BxHy(s) + O2(g) ---> (x/2)B2O3(s) + (y/2)H2O(g)

If 0.148 g BxHy yields 0.422 g B2O3 when burned in excess O2, what is the empirical formula of BxHy?

Solution:

1) Grams of B in 0.422 g of B2O3:

0.422 g times (21.622 / 69.619) = 0.131 g <--- also grams of B in BxHy

2) Grams of H in 0.148 g of BxHy:

0.148 − 0.131 = 0.017 g

3) Determine moles of each:

moles B in 0.131 g ---> 0.01212963
moles H in 0.017 g ---> 0.01686508

4) Look for smallest whole-number ratio:

B ---> 0.01212963 / 0.01212963 = 1
H ---> 0.01686508 / 0.01212963 = 1.4

Multiply by 5 gives 5 and 7

B5H7