Moles

Question 1

Relative Atomic Mass (Ar)

Answer 1

The relative atomic mass of an element is the weighted average mass of the isotopes of the element.

To work it out:

Isotopic mass A x % abundance/ number of atoms. + isotopic mass B x % abundance/number of atoms

Question 2

Relative formula mass (Mr) (RFM)

Answer 2

The relative formula mass of a compound is the total of the relative atomic masses added up in the ratio shown in the chemical formula

For example, what is the RFM of MgO

1. We find the mass numbers of both elements (Mg = 24 and O = 16)
2. We find how many atoms there are which is shown in the question ( 1 atom of Mg and 1 atom of O)
3. Hence, the RFM of MgO = 24 + 16 = 40

For example, what is the RFM of MgF2

1. We find out the mass numbers of the elements (Mg = 24 and F =19)
2. We find out how many number of atoms there are (Mg = 1 and F = 2)
3. Hence the RFM is 24 + (19x2) = 62

Question 3

Percentage composition

Answer 3

Percentage composition = n x (Ar of element) / Mr of compound x 100
Where n is the number of atoms

For example, sodium chloride NaCl. Find the percentage composition

REMEMBER: The percentages should total to 100%

Atomic mass of Na = 23
Atomic mass of Cl = 35.5
RFM = 23 + 35.5 = 58.5

Na% = 1 x 23/ 58.5 x 100 = 39.3%
Cl % = 100 - 39.3 = 60.68%

Question 4

Definition of moles

Answer 4

The amount of substance in the relative atomic or formula mass of a substance is known as one mole of that substance

The units is mol

Question 5

Equation and triangle to work out the number of moles

Answer 5

Number of moles = mass (g) / RFM

Mass at the top, left is moles and right is RFM for triangle

Question 6

Definition of empirical formula and molecular formula

Answer 6

Empirical formula is the simplest whole number ratio of elements in a compound.

Molecular formula is the ratio of atoms and elements in a compound.

For example, Butene’s molecular formula is C4H8. What is the empirical formula?

In order to find it, we have to try to find a common factor between the numbers in order to simplify it. Here, the common factor is 4 and so the empirical formula is therefore CH2.

Question 7

Using moles to calculate empirical formula

Answer 7

For example, a compound containing 1.4g of nitrogen and 0.3g of hydrogen. Find the formula.

Steps:

1. Fill in the mass from the question
2. Find the atomic mass from the Periodic table
3. Calculate the number of moles using the equation (number of moles = mass/atomic mass)
4. Calculate the mole ration by dividing by smallest number

Mass - N: 1.4 and H: 0.3
Atomic mass - N: 14 and H: 1
Moles - N: 0.1 and H: 0.3
Mole ratio - N: 0.1/0.1 = 1 and H: 0.3/0.1 = 3
Therefore the formula is NH3

Question 8

Working out the empirical formula with percentages

Answer 8

For example, a compound contains 75% carbon and 25% hydrogen. Find the empirical formula

The percentages give us the mass per 100g

Mass in 100g
Carbon - 75
Hydrogen - 25

Relative atomic mass
Carbon - 12
Hydrogen - 1

Moles
Carbon - 75/12 = 6.25
Hydrogen - 25/1 = 25

Ratio
Carbon - 6.25/6.25 = 1
Hydrogen - 25/6.25 = 4

Therefore the empirical formula is CH4

If the ratio ends up being a decimal, you have to times the ratio by a number so that it becomes a whole number. For example, if the ratio for carbon was 1.5, you times by 2 in order to get it to a whole number. HOWEVER, you need to times ALL of the ratios by that number and so you times the hydrogen also by 2. YOU CAN’T JUST ROUND IT UP!