Empirical Formulae

Question 1

what does an empirical formula show

Answer 1

the smallest whole number ratio of atoms of each element in a compound

Question 2

what is a simple experiment that is done in order to calculate the empirical formula of copper oxide for example

Answer 2

• place a known mass of copper oxide into a tube
• heat it in a stream of hydrogen gas or natural gas
• burn off excess gas at the end of the experiment for safety reasons
• after it has cooled, remove and weigh the solid copper

Question 3

what happens when you heat the hydrogen gas with the copper oxide

Answer 3

• the hydrogen atoms in the gas react with the oxygen atoms in the copper oxide to form steam
• resulting in the formation of water and solid copper

Question 4

what would be observed during the experiment

Answer 4

• the solid gradually changes colour from black to orange-brown
• which is the colour of copper

Question 5

what could be done to make sure you have more accurate masses of copper

Answer 5

• heat the solid again in the stream of gas
• this is to check whether its mass changes any more
• as heating to a constant temp suggests the conversion to copper is complete

Question 6

what is the order in which you should write the values for calculating the empirical formula of a substance

Answer 6

• mass
• Mr
• mass / Mr (moles)
• ratio
• empirical formula

Question 7

for the experiment, the mass of the copper oxide was 4.28g and the mass of copper was 3.43g. how would you calculate the empirical formula for it

Answer 7

• 4.28 - 3.43 = 0.85g of O
• mass: Cu = 3.43, O = 0.85
• Mr: Cu = 63.5, O = 16
• mass / Mr: = Cu = 0.054, O = 0.0531
• ratio: 1 : 1
• EF: CuO

Question 8

how would you calculate the empirical formula of a compound composed of 38.4% C, 4.8% H and 56.8% Cl

Answer 8

• per: C = 38.4, H = 4.8, Cl = 56.8
• Mr: C = 12, H = 1, Cl = 35.5
• per / Mr: C = 3.2, H = 4.8, Cl = 1.6
• ratio: 2 : 3 : 1
• EF = C2H3Cl

Question 9

A 1.87g sample of an organic compound was completely burned, forming 2.65g of carbon dioxide and 1.63g of water. how would you firstly calculate the masses of carbon and hydrogen that form the organic compound and therefore the mass of the organic compound

Answer 9

• you would need to calculate the masses of carbon and hydrogen in the carbon dioxide and water
• Mr of CO2 is 44, but the proportion of C in that is 12 / 44
• Mr of H20 is 18, but proportion of H in that is 2 / 18
• mass of carbon = (2.65 x 12) / 44 = 0.723g
• mass of hydrogen = (1.63 x 2) / 18 = 0.181g
• mass og hydrocarbon = 0.904g

Question 10

why is this calculation only possible if the organic compound completely burns

Answer 10

• because you assume that all of the carbon in the carbon dioxide comes from the organic compound
• and all of the hydrogen in the water comes from the organic compound

Question 11

how would you then calculate the mass of oxygen

Answer 11

1.87 - 0.904 = 0.996g

Question 12

what is the empirical formula of the organic formula including oxygen

Answer 12

• mass…
• Mr…
• mass / Mr…
  ratio: C = 1, H =3, 0 = 1
• EF = CH3O