Percent Composition, Empirical, and Molecular Formulas Lesson Objectives Calculate percent composition from a chemical formula Determine the empirical and molecular formulas for a substance This presentation will teach you to calculate the percent composition by mass of the elements in a chemical formula and how to determine the empirical and molecular formulas for a substance.

Percent Composition Percent composition describes the contribution of each element to the total mass of a compound Calculations done for one mole of a substance Ratio of one part of the compound to the whole compound The percent composition describes the contribution of each element to the total mass of the compound. The calculations are typically done for one mole of a substance since the molar mass is a relatively easy value to calculate. You can think of the percent composition as the ratio of one part of the compound to the whole compound. The formula for percent composition can be expressed as the mass of the part divided by the mass of the whole multiplied by 100 (since it is a percentage). Since we will be calculating the percent composition of an element in one mole of a compound. You can think of the formula more specifically as the mass of the element in one mole of the compound divided by the molar mass of the whole compound, multiplied by 100.

Ex) Find the percent composition of CaCl2 Calculating Percent Composition To determine the percent composition, you need to know Formula of the compound Molar mass of each element in the compound Molar mass of the compound Ex) Find the percent composition of CaCl2 In order to determine the percent composition of a substance, you need to know the formula of the compound, the molar mass of each element in the compound, and the molar mass of the entire compound. For example, let’s find the percent composition of both calcium and chlorine in calcium chloride, CaCl2. First, let’s find the percent composition of calcium. The percent composition of calcium will equal the mass of calcium in 1 mol CaCl2 divided by the molar mass of calcium chloride. There is 1 mol Ca in 1 mol CaCl2. Therefore, the mass of calcium in 1 mol CaCl2 is 40.078 g. The molar mass of calcium chloride is calculated by adding the mass of one mole of calcium (40.078 g) to the mass of two moles of chlorine (each with a mass of 35.453 g). The total molar mass of CaCl2 is 110.984 g/mol. Now the percent composition can be determined. The mass of calcium in 1 mol CaCl2, 40.078 g, divided by the molar mass of calcium chloride, 110.984 g/mol, times 100, gives us the percentage of calcium in calcium chloride, 36.111%. Following a similar process, we’ll calculate the percent composition of chlorine. It is important to notice that there are two moles of chlorine in every mole of the compound so we must multiply the mass of chlorine, 35.453 by two, to find the mass of chlorine in 1 mol CaCl2, 70.906 g. We will divide this by the molar mass of the whole compound, 110.984, and multiply by 100 to get the percent composition for chlorine, 63.889%. This answer is reasonable since the masses of calcium and chlorine were similar and there is twice as much chlorine in the formula of the compound. In addition, we can check our work by making sure the percentages add up to 100. Since the sum of 36.111% and 63.889% is 100%, our answers are correct.

Empirical and Molecular Formulas Empirical formula – formula that shows the simplest whole number ratio of elements in a compound Molecular formula – formula of one molecule or formula unit of a compound as it actually exists Can be a multiple of the empirical formula Ex) N2O4 is a molecular formula with an empirical formula of NO2 Sometimes the empirical and molecular formulas are the same Ex) NH3 is the molecular formula and the empirical formula of ammonia The empirical formula of a compound is a formula that shows the simplest whole number ratio of elements in a compound. The molecular formula of a compound is the formula of one molecule or formula unit of a compound as it actually exists. The molecular formula can be a multiple of the empirical formula. For example, the formula N2O4 is a molecular formula because it shows the compound as it actually exists. It can be simplified by decreasing everything by a factor of two to get an empirical formula of NO2. Sometimes the empirical and molecular formulas of a compound are the same. For example, NH3 is the molecular formula for ammonia since an ammonia molecule actually exists as one nitrogen atom bonded to three hydrogen atoms. However, this formula is also the simplest whole number ratio between the atoms. Therefore, it is also the empirical formula. N H

Empirical Formula from Molecular Formula Reduce a molecular formula to an empirical formula by dividing the subscripts by a factor that produces the lowest whole number ratio of subscripts Ex) C6H12 Empirical formula is CH2 Ex) H2SO4 Empirical formula is the same as the molecular, H2SO4 A molecular formula can be reduced into an empirical formula if all the subscripts can be divided by the same factor to produce the lowest whole number ratio of subscripts. For example, in the formula C6H12 both 6 and 12 can be reduced by a factor of 6. This gives an empirical formula of CH2. However, in the formula H2SO4, for example, there is no way to reduce the formula since there is already only one atom of sulfur. In this case, the empirical formula is the same as the molecular formula, H2SO4.

Molecular Formula from Empirical Formula and Molar Mass Molecular formula can be a multiple of the empirical formula Determine the molar mass of the empirical formula Divide the molar mass of the molecular formula by the molar mass of the empirical formula to find the factor Multiply the empirical formula by the factor Ex) What is the molecular formula of a compound with an empirical formula of CH2O and molar mass 180.18 g/mol? You may also be asked to find the molecular formula of a compound from its empirical formula and the molar mass. Since the molecular formula can be a multiple of the empirical formula, we just need to find the factor to multiply the empirical formula by. Let’s work an example problem as we go through the steps for finding the molecular formula from the empirical formula and molar mass. What is the molecular formula of a compound with an empirical formula of CH2O and a molar mass of 180.18 g/mol? First, we need to determine the molar mass of the empirical formula. Since the formula contains one mole of carbon atoms, two moles of hydrogen atoms, and one mole of oxygen atoms, we’ll determine the molar mass by finding the sum of 12.011 plus the product of 1.0079 times two plus 15.999. Therefore, the molar mass of the empirical formula is 30.026 g/mol. Next, we’ll divide the molar mass of the molecular formula by the molar mass of the empirical formula to determine the factor we’ll multiply the empirical formula by to get the molecular formula. In this case, we’ll divide 108.18 g/mol by 30.026 g/mol and find that the difference between the empirical and molecular formulas is a factor of six. The last step is to multiply the empirical formula by the factor. When we multiply CH2O by a factor of six, it becomes C6H12O6.

Empirical and Molecular Formulas from Percent Composition and Molar Mass Percent composition gives a ratio of the masses of the elements in a compound Convert masses into moles Assume a 100 g sample Divide each mole value by the smallest mole value Whole number products are the subscripts for the empirical formula (see rounding rules) Ex) A compound is 82.8% carbon and 17.2% hydrogen. The molar mass is 58.14 g/mol. Find the empirical and molecular formulas. The empirical and molecular formulas for compounds can also be determined from the percent composition and the molar mass of the compound. Let’s work an example problem as we go through the steps. A compound is 82.8% carbon and 17.2% hydrogen. The molar mass is 58.14 g/mol. Find the empirical and molecular formulas. The percent composition gives a ratio of the masses of the elements in a compound. However, chemical formulas show mole ratios, not mass ratios. So, we’ll convert the masses into moles in order to determine a chemical formula. First, we’ll need to convert the percentages into gram amounts. To do this, we’ll assume that we have 100 grams of the compound. If we have 100 g of the compound and 82.8% of that is carbon, then we have 82.8 grams of carbon. Similarly, if we have 100 g of the compound and 17.2% of that is hydrogen, then we have 17.2 grams of hydrogen. We’ll need to convert grams of the substances into moles since chemical formulas show mole ratios between elements. We’ll multiply 82.8 grams of carbon by one mole of carbon over the number grams of carbon in a mole of carbon, 12.011, and find that it is equal to 6.89 moles of carbon. Then we’ll do the same with hydrogen. We’ll multiply 17.2 grams of hydrogen by one mole of hydrogen over the number grams of hydrogen in a mole of hydrogen, 1.0079, and find that it is equal to 17.1 moles of hydrogen. Since 6.89 moles and 17.1 moles are not simple whole number ratios, additional steps are needed to determine the empirical formula of the compound. Next, we’ll divide each mole value by the smallest mole value to determine the whole number ratios between the elements. Since the smallest value was 6.89 moles, we’ll divide both mole values by this number. When 6.89 moles of carbon is divided by 6.89, we get one mole of carbon. When 17.1 moles of hydrogen is divided by 6.89, we get about 2.48 moles of hydrogen. This is not close enough to a whole number for us to round up to three or down to two. Let’s examine the rules for rounding in situations like these. (Click the link to see rules pertaining to when to round, when not to round and what to do instead.) 2.48 is about 2.5, which we’ll multiply by two to create the whole number five. Since we multiplied the number of moles of hydrogen by two, we’ll also need to multiply the number of moles of carbon by two. This gives two moles of carbon. These whole number products are the subscripts for the empirical formula. Therefore, the empirical formula of the compound is C2H5.

Rounding in Empirical Formula Calculations If a number ends with a decimal of .01 or .99 then round to the nearest whole number Sometimes the decimals will not be close enough to a whole number to round Create whole numbers by multiplying all the mole values by the smallest possible factor Look for these decimals during calculations Ends in .5 Equivalent of ½ Multiply by 2 Ends in .33 or .66 Equivalents of or Multiply by 3 Ends in .25 or .75 Equivalents of ¼ or ¾ Multiply by 4 When performing empirical formula calculations you will not always get perfect whole numbers for mole ratios. If your numbers end with .01 or .99, then they are close enough to round to the nearest whole number. However, when decimals are not close enough to a whole number, you should NOT round to the whole number. Instead, you should create whole numbers by multiplying all of the mole values by the smallest possible factor. Look out for the following decimals during calculations: numbers that end with decimals of 0.5, the equivalent of one half, numbers that end with decimals of .33 and .66, the equivalents of one-third and two-thirds respectively, and numbers that end with decimals of .25 and .75, the equivalents of one-fourth and three-fourths respectively. When you have a mole value ending in .5, multiply everything by two to get whole numbers. When you have mole values ending in .33 or .66, multiply everything by three to get whole numbers. When you have mole values ending in .25 or .75, multiply everything by four to get whole numbers. Back to Problem

Empirical and Molecular Formulas from Percent Composition and Molar Mass Percent composition gives a ratio of the masses of the elements in a compound Convert masses into moles Assume a 100 g sample Divide each mole value by smallest mole value Whole number products are the subscripts for the empirical formula Divide the molar mass of the molecular formula by the molar mass of the empirical formula to find the factor Multiply the empirical formula by the factor Ex) A compound is 82.8% carbon and 17.2% hydrogen. The molar mass is 58.14 g/mol. Find the empirical and molecular formulas. Now that we’ve found the empirical formula, let’s find the molecular formula, just as we did before, when we were given the empirical formula and the molar mass of the compound. Here, we know that the empirical formula is C2H5 and the molar mass of molecular formula is 58.14 g/mol. As before, we’ll divide the molar mass of the molecular formula by the molar mass of the empirical formula to find the factor to multiply the empirical formula by. Before we can do that, we must calculate the molar mass of the empirical formula. Since there are two moles of carbon and five moles of hydrogen in one mole of C2H5, the sum of the product of two times 12.011 and the product of five times 1.0079, will equal the molar mass of the empirical formula, 29.062 g/mol. Now, we’ll divide the molar mass of the molecular formula, 58.14 g/mol, by the molar mass of the empirical formula, 29.062 g/mol, and find that the factor is 2. Finally, we’ll multiply the empirical formula by the factor, 2, to find the molecular formula, C4H10.

Practice Problem Analysis of a chemical used in photographic developing fluid indicates a chemical composition of 65.45% C, 5.45% H, and 29.09% O. The molar mass is found to be 110.0 g/mol. Determine the molecular formula.