Additional Chemistry: Topic - How Much?

Relative Atomic Mass

• Carbon has a relative atomic mass (Ar) of 12 as  carbon has the mass number 12. Similiarly, silver has a relative atomic mass of 108, as this is the mass number.
• Atoms of different isotopes have different masses. The relative atomic mass will be the average of the relative atomic masses of all the isotopes.
• The relative atomic mass of an element taken in grams = 1 mole

Relative Formula Mass

• The mass of a compound calculated using relative atomic masses.
• e.g. Lithium Iodide - Li (7) and I (127). 7+127=134

% Compositon by Mass

e.g. % mass of each element in potassion carbonate:

1) Find relative formula mass: K2CO3 --> (2x39)+12+(3x16) = 138 

2) %K = 2x39 / 138 x 100 = 56.5% and so on. Should add up to 100%

Simplest Empirical Formula

• It is the simplest ratio of atoms in a formula

e.g. An oxide of sodium was found to contain 4.6g of sodium and 1.6g of oxygen by mass. Find its simplest formula.

Na (23) and O (16)

--> 4.6 / 23 and 1.6 / 16

--> 0.2 and 0.1

--> divide by the smallest number - 0.1

--> = 2:1

Masses and Formulae

e.g. What mass of calcium oxide is needed to provide 2.0g of Ca

--> Relative formula mass of CaO = 40 + 16 = 56

-->40g Ca comes from 56g of CaO

--> 1g Ca comes from 56g / 40g of CaO

--> 2g Ca comes from (56 / 40) x 2 of CaO

--> = 2.8g

• When faced with 'kg' rather than 'g', conversion is not necessary as the formula refers only to units, not to 'g' specifically

Formulae and Equations

e.g. Calculate the mass of copper oxide that can be obtained from 1.235g CuCO3

--> Relative formula mass - CuCO3 (123.5) -> CuO (79.5) + CO2 (44)

--> 123.5g CuCO3 gives 79.5g CuO

--> 1g CuCO3 gives 79.5 / 123.5g CuO

--> 1.235g CuCO3 gives (79.5 / 123.5g) x 1.235

--> = 0.795g

• If it is: what mass CuCO3 needed to make (x amount) CuO, swap so that rather than 79.5 / 123.5, it is 123.5 / 79.5

Percentage Yield

% yield = amount of product that formed / amount of product that should form x 100

--> 12.5g of Zinc Carbonate were heated. 8.1g ZnO should be formed.

--> Only 3.05g were actually formed.

--> (3.05 / 8.1) x 100

--> = 37.65%

Percentage Atom Economy

% atom economy = Mr of product wanted from reaction x 100 / Mr of all products

e.g. ZnCO3 -> ZnO + CO2  (The ZnO is the wanted product)

--> Mr of ZnO = 81                   Mr of ZnCO3 = 125

--> (81 x 100) / 125

--> = 64.8%