Other Chem Math and GFMs

Aim CE7: How much oxygen is in water by mass?

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???

REALLY Calculating Percent Composition

• Using the percent composition formula on Table TPercent Composition = mass of part x 100%

mass of whole

• We can calculate the percent composition of sugar in a piece of gumo Mass of UNchewed gum = 4.00 gramso Mass of CHEWED gum = 1.30 gramso Mass of sugar in gum = 2.70 gramso Percent Composition = mass of sugar x

100% of sugar UNchewed gum mass= 2.70 grams x 100 % = 67.5 %

4.00 grams

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 atoms2.02 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 H2O

Practice – calculate the % composition in each:

1. The percent composition of hydrogen in ammonia (NH3).

2. The percent composition of iron in iron III oxide (Fe2O3)

3. The percent composition of oxygen in lead IV sulfate (Pb(SO4)2).

Types of formulas – there are different ways to represent compounds

•Chemical formulas o general term for abbreviating compoundso Example: CO2

•Structural formulaso a way of drawing the shape and form of a

compoundo Example: O=C=O

•Molecular Formulao The total number of atoms of each element in

the compound o Synonymous with the term chemical

formulas

• Empirical Formulao Represents the ratio only atom to each other

in a given substance o Determined by dividing all the elements by a

common multipleo Examples

Molecular formula Empirical formulaActual N2O4 Ratio NO2

Actual C6H12O6 Ratio CH2O

Actual P2O10 Ratio PO5

Actual H20* Ratio H20** Note – compound formulas that cannot be

divided by a common multiple are both the molecular AND empirical formula for that compound

•Empirical formula units are part of the molecular formula of a given compound

•Example: o Molecular formula = multiple x Empirical formula

N2O4 = 2 x NO2

C6H12O6 = 6 x CH2O

o This means the mass of the empirical formula x the multiple = the molecular formula mass

N2O4 = 2 x NO2

(14.0g x 2)+(16.0g x 4) = 2 x (14.0g + (2 x 16.0g))

C6H12O6 = 6 x CH2O

(12gx6)+(1gx12)+(16gx6) =6 x (12g + (1gx2) + 16.0g)

•Using the empirical formula and the molecular mass, the molecular formula can be calculated

• Example 1 – the empirical formula of a molecular substance is CH2, 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 CH2

o Molecular mass = 70.0 g = 2 empirical units

o Empirical mass = 14.0 g

therefore (CH2 x 5) = C5H10

•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) = N2O2