Combustion Analysis: Problems 11 - 20


Go to problems 1 - 10 of combustion analysis.

Go to a discussion of empirical and molecular formulas.

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Problem #11: 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 #12: 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 #13: 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.)

Go to answers for 11, 12 and 13


Problem #14: 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 x (12.011 / 44.009) = 15.985 g of C in 40.10 g sample
Hydrogen: 14.98 g x (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 sample times (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? The ChemTeam must confess that he did not figure this out on his own, but learned it from an answer to this question on Yahoo Answers Chemistry. Also, notice how the oxygen is determined by subtraction after everything else is calculated. This is the 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 #15: 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 times (12.011 / 44.009) = 0.4694 g of C
hydrogen: 1.18 g times (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 #16: 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 times (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 times (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 #17: 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 #18: 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 x (12.011 g / 44.0 g) = 1.1902 g of C
1.1902 g / 2.52 g = 47.23%

2) Hydrogen:

0.892 g x (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 by 0.7843:

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

Empirical formula is C5H5NOS


Repeat Problem #18: 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 x (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 x (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 #19: 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 times (12.011 / 44.01) = 0.0120 g
H: 0.0270 g times (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 #20: 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

Go to problems 1 - 10 of combustion analysis.

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