# Quantitative chemistry

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## Calculating percentage by mass

% by mass = Ar of element/Mr of total compound x 100

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## Moles in chemistry

A 'mole' is an amount of something. Similar to couple or dozen but value of a mole is 6.02 x 10 to the power of 23. -  Avogadro's number.

Moles: usually used to describe a number of atoms.

1 mole of an element has a mass which is equal to the Ar

1 mole of a compound has a mass which is equal to the Mr

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## The empirical formula

Empirical formula:  Simplest whole number ratio of each element in a compound.

Step 1: work out ratio amounts of atoms/ ratio of moles by:

• dividing the mass/percentage of each element by its Ar. - this calculates the number of moles it contains.

Step 2: Simplify the ratio by dividing all numbers by the smallest number.

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## Calculating the empirical formula : Example

E.g. A compound is 18.2% potassium, 59.4% iodine and 22.4% oxygen. Find the empirical formula of the compound.

Step 1 :

K  = 18.2 / 39  = 0.47

I   = 59.4 / 127 = 0.47

O = 24.4 / 16   = 1.4

Step 2:

K- 0.47 / 0.47 = 1

I- 0.47 / 0.47  = 1

O- 1.5 / 0.47  = 3

Step 3: Empirical formula = KIO3

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## Calculating reacting masses

Calculating reacting masses is calculating the ratio of reactants and products in a reaction. It is important industrially to ensure no valuable reactant is wasted or unused, otherwise the process will not be as profitable.

E.g. Calculating the maximum mass of Methanol that can be produced from 14g of Carbon monoxide:

Step 1: Balanced equation - CO + 2H2 --> CH3OH

STEP 2: calculate Mr of products and reactants:

CO = 28g    CH3OH  = 32g

*we go from 28 to 32, to calculate how we divide 32 by 28 = 1.14 and multiply 1.14 by 14g (produced from 14g of Carbon monoxide)

Step 3:

28g of CO --> 32g of CH3OH .... so .... 14g of CO ---> 16g of CH3OH

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