Chapter 3:  Composition of Compounds and Experimental Determination of Chemical Formulas

 

Section 3-1: Percent Composition by Mass

Section 3-2: Determination of Empirical and Molecular Formulas

Section 3-3: Experiment Determining the Formula of a Hydrate

Section 3-4: Experiment - Combustion Analysis

Chapter 3 Practice Exercises and Review Quizzes

 

 

 

 

 

 

Section 3-1:  Percent Composition by Mass

 

We can determine the percent by mass of a given element in a compound as follows:

 

 

For example, to determine the percent by mass of hydrogen in water, we divide the grams of hydrogen in 1 mole of water by the molar mass of water:

 

 

Sample Exercise 3A:

 

Calculate the percent by mass of carbon in (C2H5)2NH.

 

Solution:

 

 

 

 

 

Section 3-2:  Determination of Empirical and Molecular Formulas

 

A molecular formula gives the actual number of atoms of each element in one molecule of a compound.  For example, since a glucose molecule contains 6 carbon atoms, 12 hydrogen atoms, and 6 oxygen atoms, the molecular formula of glucose is C6H12O6. 

 

An empirical formula gives the simplest ratio of atoms of each element in a compound.  For example, since the simplest ratio in glucose is 1 C atom: 2 hydrogen atoms: 1 oxygen atom, the empirical formula of glucose is C1H2O1 or simply CH2O.  An empirical formula also gives the simplest ratio of moles of each element in a compound.  Note, however, that an empirical formula does NOT give the simplest ratio of grams of each element in a compound.

 

The percent composition by mass of an unknown compound can be determined by a variety of experimental methods, after which the empirical formula of the unknown compound can be determined as demonstrated in the problem below: 

 

Sample Exercise 3B:

 

An unknown compound was found to be 55.77% carbon and 11.70% hydrogen by mass, with the remainder being nitrogen.  Determine the empirical formula of the compound.

 

Solution:

 

First, subtract to obtain the percent by mass of nitrogen in the compound: 

 

100% – 55.77% C – 11.70% H = 32.53% N

 

Next, assume you have exactly one hundred grams of the unknown compound, in which case the percent by mass of each element becomes equal to the grams of each element, and then convert grams of each element to moles:

 

 

Recalling from the definition above that an empirical formula gives the simplest ratio of moles of each element, divide the moles of each element by the smallest number of moles in the group and then round each result to the nearest whole number:

 

4.644 mol C: 11.61 mol H: 2.322 mol N (divide each by 2.322)

= 2 mol C: 5 mol H: 1 mol N

 

Therefore, the empirical formula of the unknown compound is C2H5N.

 

The molar mass of a compound can be determined by a variety of experimental methods.  If both the empirical formula and the molar mass of a compound are known, the molecular formula of the compound can be determined as demonstrated in the following problem:

 

Sample Exercise 3C:

 

A compound with empirical formula C2H5N is found to have a molar mass of about 86 g/mol.  What is the molecular formula of the compound?

 

Solution:

 

First, calculate the molar mass of the empirical formula (EM):

 

EM = 2(12.01) + 5(1.008) + 14.01 = 43.07 g/mol

   

Next, divide the molar mass of the compound given in the problem by the calculated molar mass of the empirical formula and round the result to the nearest whole number:

 

Finally, multiply the subscripts in the empirical formula by the whole number above to obtain the molecular formula:

 

C2x2H5x2N1x2 = C4H10N2 = molecular formula

 

 

 

 

Section 3-3:  Experiment – Determining the Formula of a Hydrate

 

A hydrate is an ionic compound with water incorporated into the solid crystal.  For example, the hydrate MgSO47H2O contains 7 moles of water for every mole of MgSO4 in the crystal.  The formula of a hydrate can be determined experimentally by heating a known mass of the hydrate until all the water is removed and then finding the mass of the remaining anhydrous compound.  The mass of water removed can be obtained by subtraction, after which the calculated ratio of moles of water to moles of the anhydrous compound yields the formula of the hydrate, as demonstrated in the following problem:

 

Sample Exercise 3D:

 

To determine the value of x in the formula of the hydrate BaCl2xH2O, a student heated a sample of the hydrate in an evaporating dish over a Bunsen burner to remove all the water and recorded the following data:

 

Mass of Empty Evaporating Dish

22.36 g

Mass of Evaporating Dish + Hydrate

32.23 g

Mass of Evaporating Dish + Anhydrous BaCl2

30.20 g

 

Determine the value of x to the correct number of significant figures and the most likely formula of the hydrate.

Solution:

 

First, subtract the mass of the evaporating dish to find the mass of the hydrate before heating and the mass of the anhydrous BaCl2 after heating:

 

32.23 g – 22.36 g = 9.87 g hydrate before heating

30.20 g – 22.36 g = 7.84 g anhydrous BaCl2 after heating

 

Next, subtract the results above to find the mass of water removed by heating:

 

9.87 g – 7.84 g = 2.03 g water removed

 

Note that we keep 2 decimal places in all the masses calculated by subtraction above because all the original measurements had 2 decimal places.

 

Finally, convert the masses of water and BaCl2 to moles and find the ratio of moles of water to moles of BaCl2:

 

M of H2O = 2(1.008) + 16.00 = 18.02 g/mol

M of BaCl2 = 137.3 + 2(35.45) = 208.2 g/mol

 

Note that we will keep an extra sig. fig. (4 instead of 3) in the intermediate grams to moles calculations below, but then round our final ratio to the correct number of sig. fig.s.

 

 

 

Based on the ratio calculated above, there are most likely 3 moles of water for every mole of BaCl2 in the hydrate.  Therefore, the most likely formula of the hydrate is BaCl23H2O. 

 

Note that the calculated ratio in hydrate experiments may not be a whole number due to experimental error.  For example, if the hydrate is underheated and not all the water is removed, the final mass recorded will be too high.  As a result, the mass of the anhydrous compound found by subtraction will be too high and the mass of water found by subtraction will be too low, leading to an erroneously low ratio of moles of water to moles of anhydrous compound.

 

 

 

Section 3-4:  Experiment – Combustion Analysis

 

In combustion analysis, the empirical formula of a compound can be determined by completely burning a known mass of the compound in the presence of oxygen gas and then finding the masses of certain products.  When a compound with the general formula CxHyQz burns completely in the presence of oxygen gas, the reactants and products of the combustion reaction are:

 

CxHyQz + O2 CO2 + H2O + other product (note:  no other product if element Q = oxygen or if z = 0)

 

During the combustion reaction above, all the carbon atoms in the compound burned will end up in the CO2 produced, and all the hydrogen atoms in the compound burned will end up in the H2O produced.  Therefore, the moles and mass of carbon in the CO2 will be equal to the moles and mass of carbon in the compound burned, and the moles and mass of hydrogen in the H2O will be equal to the moles and mass of hydrogen in the compound burned.  As demonstrated below, knowing the masses of carbon and hydrogen in the compound burned allows us determine the mass of element Q in the compound by subtraction, after which we can convert the mass of element Q to moles in order to determine the empirical formula of the compound:

 

Sample Exercise 3E:

 

(a) An unknown compound contains carbon, hydrogen, and oxygen.  When 6.75 grams of the compound was burned completely in the presence of oxygen gas, 9.89 grams of CO2 and 4.05 grams of H2O were produced.  Determine the empirical formula of the compound.

(b) In a separate experiment, the molar mass of the unknown compound was found to be about 60 g/mol.  Determine the molecular formula of the compound.

 

Solution:


 

 

 

Chapter 3 Practice Exercises and Review Quizzes:

 

3-1) Calculate the percent by mass of oxygen in Ca3(PO4)2.

Click for Solution

 

3-1)

 

 

 

 

3-2) What is the empirical formula of C6H12O3?

Click for Solution

 

3-2) divide all subscripts by 3 to get simplest ratio = C2H4O = empirical formula

 

 

 

3-3) (a) An unknown compound was found to be 26.17% chlorine and 56.10% fluorine by mass, with the remainder being carbon.  Determine the empirical formula of the compound.

       (b) In a separate experiment, the molar mass of the unknown compound was found to be about 271 g/mol.  Determine the molecular formula of the compound.

Click for Solution

 

3-3) (a) 100% - 26.17% Cl – 56.10% F = 17.73% C by mass

 

Assume one hundred grams of unknown compound:

 

 

 

 

1.476 mol C: 0.7382 mol Cl: 2.953 mol F (divide each by 0.7382)

= 2 mol C: 1 mol Cl: 4 mol F

 

Therefore, empirical formula is C2ClF4.

 

(b) EM = 2(12.01) + 35.45 + 4(19.00) = 135.5 g/mol

 

 

C2x2Cl1x2F4x2 = C4Cl2F8 = molecular formula

 

 

 

3-4) To determine the value of x in the formula of the hydrate CaSO4xH2O, a student heated a sample of the hydrate in an evaporating dish over a Bunsen burner to remove all the water and recorded the following data:

 

Mass of Empty Evaporating Dish

19.73 g

Mass of Evaporating Dish + Hydrate

21.50 g

Mass of Evaporating Dish + Anhydrous CaSO4

20.80 g

 

Determine the value of x to the correct number of significant figures and the most likely formula of the hydrate.

Click for Solution

3-4)

21.50 g – 19.73 g = 1.77 g hydrate before heating

20.80 g – 19.73 g = 1.07 g anhydrous CaSO4 after heating

 

1.77 g – 1.07 g = 0.70 g water removed

 

M of CaSO4 = 40.08 + 32.07 + 4(16.00) = 136.2 g/mol

 

 

 

Based on the ratio calculated above, there are most likely 5 moles of water for every mole of CaSO4 in the hydrate.  Therefore, the most likely formula of the hydrate is CaSO45H2O. 

 

 

 

3-5) (a) An unknown compound contains carbon, hydrogen, and sulfur.  When 9.88 grams of the compound was burned completely in the presence of oxygen gas, 18.3 grams of CO2 and 3.76 grams of H2O were produced.  Determine the empirical formula of the compound.

       (b) In a separate experiment, the molar mass of the unknown compound was found to be about 142 g/mol.  Determine the molecular formula of the compound.

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3-5)

 

 

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