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**Section 7.3 - Percent Composition and Chemical Formulas**

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**Calculating the Percent Composition of a compound**

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The relative amounts of each element (Part) in a compound (Whole) are expressed as the Percent Composition or the Percent by Mass of each element in the compound.

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The percent composition of a compound has as many percent values as there are different elements in the compound.

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**The percent composition of CaCO3 is:**

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To calculate the percent composition of a compound, use the chemical formula to determine which elements and how many moles of each are in the compound.

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Divide the total mass of one of the elements in the compound by the molar mass of the compound then multiply by 100 to convert to a percent.

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Remember that the total mass of the element is the molar mass of the element multiplied by its subscript in the formula.

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**Example: What is the percent composition of FeSO4?**

= 36.7 % S = = 21.1 % O = = 42.2 %

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The Percent By Mass in a compound is the number of grams of the element divided by the number of grams of the compound multiplied by 100.

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Sample Problem p 189 An 8.20 g piece of magnesium combines with 5.40 g of oxygen to form a compound. What is the Percent Composition (Percent By Mass) of this compound?

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% Mg = % Mg = % O = % O =

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**Using Percent Composition as a Conversion Factor**

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**You can use the percent composition to determine the mass of each element in a sample of a compound.**

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**Example: Aluminum makes up 52. 9 % of Al2O3**

Example: Aluminum makes up % of Al2O3. Calculate the mass of Aluminum in a 50.0 g sample of Al2O3.

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**We first assume that we have a 100. 0g sample of Al2O3. 52. 9 % of 100**

We first assume that we have a 100.0g sample of Al2O % of g is 52.9 g. We can use g divided by g sample as a conversion factor.

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**Calculating Empirical Formula**

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Empirical Formula

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EMPIRICAL FORMULA An empirical formula is the smallest whole number ratio of atoms in a formula.

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EMPIRICAL FORMULA If the atoms are in a small whole number ratio, then the moles of atoms of each element in the substance will also be in a small whole number ratio.

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**Two Sources of Data From Which to Calculate the Empirical Formula**

Experimental Analysis - What is the empirical formula of a compound if a 2.50 gram sample contains g of calcium and 1.60 g of chlorine?

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**Two Sources of Data From Which to Calculate the Empirical Formula**

Percentage Composition - A compound has a percentage composition of 40.0% C, 6.71% H, and 53.3% O. What is the empirical Formula?

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**The following is an Example of the Empirical Formula Calculation from Experimental Analysis.**

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**What is the empirical formula for a compound if a 2**

What is the empirical formula for a compound if a 2.50 g sample contains g of calcium and 1.60 g of chlorine? Determine the number of moles of each element in the compound.

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moles Ca moles Cl

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**The next step is to determine the smallest whole number ratio.**

To obtain the smallest whole number ratio, divide both numbers of moles by the smaller number of moles.

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**The mole ratio is 1 mole of Calcium to 2 moles of Chlorine.**

The Empirical formula is therefore CaCl2.

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**Example of Empirical Formula Calculation From Percentage Composition:**

A compound has a percentage composition of 40.0% C, 6.71% H, and 53.3% O. What is the empirical formula?

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**Example of Empirical Formula Calculation From Percentage Composition**

To calculate the ratio of moles of these elements, we assume a 100 g sample % of 100 g is 40.0 g.

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**Example of Empirical Formula Calculation From Percentage Composition**

Next we calculate the number of moles of each element present.

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**Example of Empirical Formula Calculation From Percentage Composition**

Mol C 3.33 Mol H 6.67 Mol O 3.33

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**Determine the smallest Whole Number Ratio.**

The number of moles of each element calculated in step 2 were: 3.33 mol C, 6.64 mol H, and 3.33 mol O. Dividing by the smallest value we obtain the smallest whole number ratio of moles.

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For C For H For O

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**With a ratio of 1-C to 2-H to 1-O**

The empirical formula is therefore --- CH2O.

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**What is the empirical formula for a compound if a 40**

What is the empirical formula for a compound if a 40.0 g sample contains 29.7 g of sodium and 10.3 g of oxygen?

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Solution: Step 1 Determine the number of moles of each element.

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**This will give the smallest whole number ratio.**

Step 2 Divide each number of moles by the smallest. This will give the smallest whole number ratio.

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**The empirical formula is therefore Na2O.**

This ratio shows that there are 2 moles of sodium for one mole of oxygen. The empirical formula is therefore Na2O.

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**Calculating Molecular Formula**

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**The Empirical Formula tells us the ratio of each kind of atom present in a compound.**

The Molecular Formula tells us the actual number of each kind of atom in a molecule of a compound.

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**If the empirical formula and the molar mass are known, the molecular formula can be calculated.**

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Example: A certain gas, having a molar mass of 78 g, has an empirical formula of CH. What is the molecular formula?

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If we multiply the mass of the empirical formula by an integer, the product must equal the molar mass.

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**Where x represents an integer.**

(CH)x = 78 g Where x represents an integer. (13)x = 78 g X = 6

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By multiplying each subscript in the empirical formula by the integer (6), we obtain the molecular formula. (CH)6 gives C6H6

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Problem: The empirical formula of a compound is CHOCl and the molar mass is 129 g. What is the molecular formula?

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