 # Mass Conservation in Chemical Reactions Mass and atoms are conserved in every chemical reaction. Molecules, formula units, moles and volumes are not always.

## Presentation on theme: "Mass Conservation in Chemical Reactions Mass and atoms are conserved in every chemical reaction. Molecules, formula units, moles and volumes are not always."— Presentation transcript:

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.

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.

Example: Empirical Formulas From Mass What is the empirical formula for a compound if a 2.50 g sample contains 0.900 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/ 40.08 g/mol = 0.0225 mol Ca) (1.60g Cl/ 35.45 g/mol = 0.0451 mol Cl) 2.To get the simplest ratio, divide both numbers of moles by the smaller one. (0.0225 mol Ca/0.0225 mol = 1 Ca) (0.0451 mol Cl/0.0225 mol = 2 Cl) 3. This calculation shows that for each mole of calcium, there are 2 moles of chlorine. 4. The empirical formula is CaCl 2

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: 1.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. 2.Next change the quantities to moles. 40.0g C/ 12.01 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. 3.Dividing each result by smallest moles (3.33), we get 1:1.99:1. This makes the empirical formula CH 2 O

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 + O 2  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. C 2 H 4 + 3O 2  2CO 2 + 2H 2 O

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.

Example: Molecular Formula Earlier, we found that the empirical formula of a compound was CH 2 O. 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 (CH 2 O). 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 C 6 H 12 O 6 REMEMBER: EF is the Simplest, Whole Number Ratio!!

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