Additional Chemistry: Topic - How Much?

Chemistry C2 Topic 3 - How Much? Revision cards for the calculations and forumulae.

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Relative Atomic & Formula Mass

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
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% Composition by Mass

% 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%

 

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(Simplest) Empirical Formula

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

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Masses and Formulae

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
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Formulae and Equations

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
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Percentage Yield

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%

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Percentage Atom Economy

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%

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Comments

sam

Thanks! This helped me greatly :)

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