1. Other Chem Math and GFMs Aim CE7: How much oxygen is in water by mass?

2. Percent Composition – the percent by mass of the parts found in a sample Percent composition looks at how much of each part makes up a given sample. In the food industry, we see this all the time… not always CORRECTLY… The question here: what is the OTHER 73% if its NOT juice???

3. REALLY Calculating Percent Composition Using the percent composition formula on Table T Percent Composition = mass of part x 100% mass of whole We can calculate the percent composition of sugar in a piece of gum o Mass of UNchewed gum = 4.00 grams o Mass of CHEWED gum = 1.30 grams o Mass of sugar in gum = 2.70 grams o Percent Composition = mass of sugar x 100% of sugar UNchewed gum mass = 2.70 grams x 100 % = 67.5 % 4.00 grams

4. REALLY Calculating Percent Composition Earliest research by scientists like John Dalton looked at percent composition. By taking apart compounds, they were able to determine the amounts of each element in them. Example: what is the percent composition by mass of hydrogen in water? Step 1 – what is the mass of H in water? 2 moles of H atoms x 1.01 grams/mole = 2.02 g Step 2 - What is the mass of 1 mole of water? Mass of 2 moles of H atoms + 1 mole of O atoms 2.02 g of H + 16.0 g of O = 18.0 g of water Step 3 – Calculate the % composition % Composition = 2.02 grams H x 100% = 11.2% 18.0 grams H 2 O

5. Practice – calculate the % composition in each: 1.The percent composition of hydrogen in ammonia (NH 3 ). 2.The percent composition of iron in iron III oxide (Fe 2 O 3 ) 3.The percent composition of oxygen in lead IV sulfate (Pb(SO 4 ) 2 ).

6. Types of formulas – there are different ways to represent compounds Chemical formulas o general term for abbreviating compounds o Example: CO 2 Structural formulas o a way of drawing the shape and form of a compound o Example: O=C=O Molecular Formula o The total number of atoms of each element in the compound o Synonymous with the term chemical formulas

7. Empirical Formula o Represents the ratio only atom to each other in a given substance o Determined by dividing all the elements by a common multiple o Examples Molecular formulaEmpirical formula Actual N 2 O 4 Ratio NO 2 Actual C 6 H 12 O 6 Ratio CH 2 O Actual P 2 O 10 Ratio PO 5 Actual H 2 0*Ratio H 2 0* * Note – compound formulas that cannot be divided by a common multiple are both the molecular AND empirical formula for that compound

8. Empirical formula units are part of the molecular formula of a given compound Example: o Molecular formula = multiple x Empirical formula N 2 O 4 = 2 x NO 2 C 6 H 12 O 6 =6 x CH 2 O o This means the mass of the empirical formula x the multiple = the molecular formula mass N 2 O 4 = 2 x NO 2 (14.0g x 2)+(16.0g x 4)= 2 x (14.0g + (2 x 16.0g)) C 6 H 12 O 6 = 6 x CH 2 O (12gx6)+(1gx12)+(16gx6)=6 x (12g + (1gx2) + 16.0g)

9. Using the empirical formula and the molecular mass, the molecular formula can be calculated Example 1 – the empirical formula of a molecular substance is CH 2, and the molar mass is 70.05 grams/mole. What is the molecular formula of the substance? o Empirical formula mass = 12.0g + (1.01g x 2) = 14.0 g per CH 2 o Molecular mass = 70.0 g = 2 empirical units o Empirical mass = 14.0 g therefore (CH 2 x 5) = C 5 H 10

10. Example 2: the empirical formula of a molecular substance is NO, and the molar mass is 60.0 grams/mole. What is the molecular formula of the substance? o Empirical formula mass = 14.0g + 16.0g = 30.0 g per NO o Molecular mass = 60.0 g = 2 empirical Empirical mass = 30.0 g units therefore (NO x 2) = N 2 O 2