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Percentage Composition

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Presentation on theme: "Percentage Composition"— Presentation transcript:

1 Percentage Composition
The percentage of the total mass of a compound contributed by an element is the percentage of that element in the compound. Examples: Elements: Copper: Copper is 100% copper because it is a single element. Compounds: Sodium chloride: Salt is a composition of two elements, chlorine and sodium.

2 Example: Salt (NaCl) We know that salt is NaCl and will always combine this way. They are always present in this ratio by mass. The ratio that they are calculated is the ratio of their atomic masses. The percentage of sodium in any sample of sodium chloride would be the atomic mass of the element divided by the formula mass and multiplied by 100. Mass Na x & Mass Cl x 100 Mass NaCl Mass NaCl (39.3% Na, 60.7% Cl) Determine formula mass (39.4% sodium 60.6% chlorine)

3 Example: Aluminum Sulfate
Find the percentage composition of aluminum sulfate. Solving Process: The formula for aluminum sulfate is Al2(SO4)3 The molecular mass is 2 Al atoms 2 x = 54.0 g mol-1 3 S atoms 3 x = 96.3 g mol-1 12 O atoms 12 x 16.0 = 192 g mol-1 342.3 g mol-1

4 (continued) Formula mass = 342.3 g mol-1
The percentage of Al: mass 2 Al____ x 100 mass of Al2(SO4)3 The percentage of S: mass 3 S____ x 100 The percentage of O: mass 12 O____ x 100 (15.8% Al, 28.2 % S, 56.1% O)

5 Mass Conservation in Chemical Reactions
Mass and atoms are conserved in every chemical reaction. Molecules, formula units, moles and volumes are not always conserved every time. Chemical reactions need to be balanced to satisfy the law of conservation of mass.

6 Empirical Formulas Empirical Formula is the SIMPLEST whole # ratio of the elements present in the compound. Using experimental data, we can find the empirical formula for a substance. We only need to know the mass (or percent) of each element in the laboratory sample. Elements in compounds combine in simple whole number ratios, such as 1:1, 1:2, 2:3, etc.

7 Example: Empirical Formulas From Mass
What is the empirical formula for a compound if a 2.50 g sample contains g of calcium and 1.60 g of chlorine? Solving Process: 1. We must determine the number of moles of each element in the compound. (0.900g Ca/ g/mol = mol Ca) (1.60g Cl/ g/mol = mol Cl) To get the simplest ratio, divide both numbers of moles by the smaller one. ( mol Ca/ mol = 1 Ca) ( mol Cl/ mol = 2 Cl) 3. This calculation shows that for each mole of calcium, there are 2 moles of chlorine. 4. The empirical formula is CaCl2

8 Example: Empirical Formulas From % Composition
A compound has a percent composition of 40% carbon, 6.71% hydrogen and 53.3% oxygen. What is the empirical formula? Solving Process: To calculate the ratio of moles of these elements, we assume an amount of the compound (100g). This makes the percentages of the compound the same as the mass in grams. So we would have 40.0 g of carbon, 6.71 g of Hydrogen and 53.3 g of oxygen in a 100 g sample. Next change the quantities to moles. 40.0g C/ g/mol = 3.33 mol of carbon 6.71g H/ 1.01 g/mol = 6.64 mole of hydrogen 53.3g O/16.0 g/mol = 3.33 mol of oxygen. Dividing each result by smallest moles (3.33), we get 1:1.99:1. This makes the empirical formula CH2O

9 Example: Empirical Formulas From Other Experimental Methods
The empirical formula can be found by direct determination. Converting a massed sample of one element to a compound (to find the mass of the second that combined with the first). Ex. 2Mg + O2  2 MgO Organic compounds are found by burning a known mass of the compound in excess oxygen, and then finding the mass of both carbon dioxide and water formed. Ex. C2H O2  2CO2 + 2H2O

10 Molecular Formulas To move from an empirical formula (which is the simplest form) to a molecular formula, we only need one more piece of information (the molecular mass). The molecular formula shows the actual number of atoms of each element in the compound, as well as the ratio of atoms.

11 Example: Molecular Formula
Earlier, we found that the empirical formula of a compound was CH2O. If we know the molecular mass of the compound is 180 g/mol, then we can determine the molecular formula. We can calculate the mass of the empirical formula (CH2O). It is 30.0 g/mol Next, we divide our molecular mass of the true formula by the molecular mass of the empirical formula (180g / 30g) = 6. This is our “multiplier”. It will take six of these E.F. units to equal 180 or one molecular formula. (Mulitplier x E.F. = M. F.) The molecular formula is C6H12O6

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