# Chemical Calculations

- Created by: zoolouise
- Created on: 19-03-16 20:49

**Chemical Calculations**

**Amount of a substance**

Atoms are too small to be counted indiviudally, therefore they're counted by weighting a collection of them where the mass of a particular fixed number of atoms is known. **As carbon-12 is the standard chosen for RAM, the number of atoms in exactly 12g of carbon-12 is chosen and is known as the mole. **

**Avogadros Constant = 6.02x10^3**

The amount of substance in moles (n), the mass (m) and molar mass (M) are linked by the equation:

**amount in moles (n) = mass (m) /molar mass (M)**

**Molar mass is the mass of one mole of a substance.****The Avogardro constant is the number of atoms per mole.**

**Calculating reacting masses**

If we're given the mass of reactants, we can find out the mass of products formed as long as the equation is balanced. The ratio between amounts in moles of reactants and products are called stoichiometric ratios.

Stoichiometry is the molar relationship between the amounts of reactants and products in a chemical reaction.

**Example (WJEC Textbook****)**

2Mg + O2 > 2MgO

This tells us that 2 moles of magnesium react with 1 mole of oxygen to produce 2 moles of magnesium oxide.

What mass of magnesium oxide forms if we burn 1.215g of magnesium in an excess of oxygen?

In order to calculate this: thie following route is used

- Change the mass of Mg into amount of moles (divide by the molar mass)
- Use the balanced equation to state the mole ratio of Mg:MgO, and decude the moles of MgO
- Change the amount in moles of MgO to mass (multiply by molar mass)
- Amount in moles of Mg = 1.215/24.3 = 0..050 mol
- The mole ratio from the equation is 2Mg:2MgO i.e 1:1 therefore 0.050 mol Mg gives 0.050 mol MgO
- Molar mass of MgO = 24.3 + 16 = 40.3 Mass MgO = 0.05 x 40.3 = 2.015g

**Empirical and Molecular formula**

**Empirical formula is the simplest formula showing the simplest whole number ratio of the amount of elements present****Molecular formula shows the actual number of atoms of each element persent in the molecule. It is a simple multiple of the empirical formula.**

We calculate empirical formula from a known mass or percentage composition that we're given. We therefore use three steps in the calculation:

1) Find the amount in moles of each element present (divide the number you have by the molar mass)

2) Find the ratio of the number of atoms present (divide both numbers by the smallest value you obtained from step one)

3) Convert these numbers into whole numbers (if they do not come out as a whole number, multiply them, but whatever you do to one, do to the other.…

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