Quatative Chemistry GCSE equations

Number of Moles= Mass ÷ Relative atomic mass

E.g.  How many moles of atoms are there in 2.4g of carbon?

12= relative atomic mass of Carbon and the mass=2.4g

2.4÷12=0.2moles

Theoretical Yield- Maximum mass of product expected from the reaction.

Actual Yield- The mass of product that is actually obtained from the real chemical reactions.

Percentage yield= (actual yield ÷ theoretical yield) x 100

Example-

What is the percentage yield of a reaction where the theoretical yield was 75kg but the actual yield was 68kg?

(68÷75) x 100= 90.7%

Atom Economy= (mass of desired product ÷ total mass of reactants) x100

Example- Whtais the atom economy of a reaction here the actual yield was 25 000 tonnes but the mass of the reactnts was 30 000 tonnes?(25000 x 30 000)x100= 83.3%

Mean= add all results ÷ number of results (do NOTinclude anomolies)

Range= difference between highest ad lowest results (do NOT include aomolies)

Uncertainty= Range ÷ 2

Example

Find te Mr of Sodium Chloride (NaCl) Each atom of NaCl contains 1Na atom and 1Cl atom.

Ar(Na)= 23    Ar(Cl)=35.5

So Mr(NaCl)= 23 + 35.5= 58.5

N2+3H2 --> 2NH3 if you have 20g of N2 and 5.0g of H2.What it is the limiting reactant?

RFM=2NH3= 28+6= 34

20g of N2

5.0g of H2

Using formula Mole=Mass ÷ RFM

N=20÷28=0.71

H= 5÷6= 0.83 

So Limiting reactat= Nitrogen

Lowest Number = limiting reactant

Highest Number = excess

What mass of sulfuric acid reacts with sodium hydroxide to produce 10g of sodium sulfate?

H₂SO₄ +2NaO àNa₂SO₄ + 2H₂O

Relativ formula mass

H₂SO₄ = (2x1)+32+(4x16)=98   Na₂SO₄= (2x23)+32+(4x16)=142

98g of Sulfuric Acid= 142g of  Sodium Sulfate.

For 1 gram = 98 ÷ 142             142÷142

For 10 grams= (98 ÷ 142)x10 = 6.9g          (142÷142) x10= 10g