It’s time to learn about.... Happy Friday! Please have out your work for the molar conversions on Page 4. We will have a brief review before the quiz.

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Presentation transcript:

It’s time to learn about...

Happy Friday! Please have out your work for the molar conversions on Page 4. We will have a brief review before the quiz.

Stoichiometry: Percent Composition At the conclusion of our time together, you should be able to: 1. Determine the percent composition of each of the elements in a compound

% Composition – Page 5 Is the percent by mass that each element contributes to the molar mass. 1) Find the molar mass of the compound. 2) Element Mass/Molar Mass * 100

EXAMPLE: What is the percent composition of hydrogen and oxygen in water? 1) What is the molar mass of water? 2) Element Mass/Molar Mass * 100 Try: CuBr 2

Percent Composition What is the percent carbon in C 5 H 8 NO 4 (the glutamic acid used to make MSG monosodium glutamate), a compound used to flavor foods and tenderize meats? a) 8.22 % C b) 24.3 % C c) 41.1 % C d) not listed

Chemical Formulas of Compounds Formulas give the relative numbers of atoms or moles of each element in a formula unit - always a whole number ratio (the law of definite proportions). NO 2 molecule: 2 atoms of O for every 1 atom of N 1 mole of NO 2 : 2 moles of O atoms to every 1 mole of N atoms If we know or can determine the relative number of moles of each element in a compound, we can determine a formula for the compound.

Types of Formulas – Page 5 Empirical Formula The formula of a compound that expresses the smallest whole number ratio of the atoms present. It is the simplest ratio possible. Ex – P4O6P4O6 C6H9C6H9 CH 2 OHCH 2 OH BrCl 2

To Obtain an Empirical Formula 1.Determine the mass in grams of each element present, if necessary. 2.Calculate the number of moles of each element. 3.Divide each by the smallest number of moles to obtain the simplest whole number ratio. 4. If whole numbers are not obtained in step 3, multiply through by the smallest number that will give all whole numbers

Page 6, #9 & #11

A sample of a brown gas, a major air pollutant, is found to contain 2.34 g N and 5.34g O. Determine the formula for this substance. This requires mole ratios, so convert grams to moles moles of N = 2.34 g of N g/mole = moles of N moles of O = 5.34 g g/mole = moles of O Formula:

Empirical Formula from % Composition A substance has the following composition by mass: % Na ; % B ; % H What is the empirical formula of the substance? Consider a sample size of 100 grams This would contain g of Na, grams of B and grams H Determine the number of moles of each Determine the simplest whole number ratio

Empirical Formula from % Composition Determine the number of moles of each 2.64 mol Na 2.65 mol B mol H Determine the simplest whole number ratio NaBH 4

Molecular Formula – Page 7 Molecular Formula The formula that states the actual number of each kind of atom found in one molecule of the compound. 1) Molar Mass/Empirical Mass 2) Empirical Formula * Answer

Molecular Formula, Page 7 #1

Calculation of the Molecular Formula A compound has an empirical formula of NO 2. The colorless liquid, used in rocket engines has a molar mass of 92.0 g/mole. What is the molecular formula of this substance? Empirical Mass = = 46.01g/mol 92.0 g/mol g/mol = ~2 N2O4N2O4

Stoichiometry: Percent Composition Let’s see if you can: 1. Determine the percent composition of each of the elements in a compound 2. Use percent compositions of a compound to determine the empirical and/or molecular formula

Empirical Formula from % Composition A substance has the following composition by mass: 12.8% C, 2.1% H, and 85.1% Br (by mass). Calculate the empirical formula and the molecular formula of this compound given that the molar mass is 188 g/mol. Consider a sample size of 100 grams Determine the number of moles of each Determine the simplest whole number ratio

Empirical Formula from % Composition Determine the number of moles of each 12.8 g/12.01 g/mol = 1.07 mol C 2.1 g/1.01 g/mol = 2.08 mol H 85.1 g/79.90 g/mol = 1.07 mol Br CH 2 Br

Molecular Formula from Empirical Formula CH 2 Br Empirical Molar Mass = g/mol Molecular Molar Mass = 188 g/mol g/mol ~2 C 2 H 4 Br 2