Empirical formula and reactant mass Learning Objective: To be able to perform calculation without a problem!

Reacting Masses

Calculating reacting masses What mass of iron could be produced from 80g of iron oxide? Fe 2 O 3 + 3CO  2Fe + 3CO 2 (56x2)+(16x3)  2x g of iron oxide would make 56g of iron ________ The balanced equation Data RAM Fe 56 O 16 Data for calculating RMM’s Simple proportion- 80 is half of 160 Calculate the formula masses of iron oxide and iron 2 Calculator

Calculating reacting masses What mass of water could be made by burning 320g of methane? CH 4 + 2O 2  2H 2 O + CO 2 12+(4x1)  2x(2+16) 36 2x g of methane would make 720g of water ________ The balanced equation Data RAM C 12 O 16 H 1 Data for calculating RMM’s Simple proportion- Calculate the formula masses of methane and water 32  72

Calculating reacting masses What mass of ethanol could be made by fermenting 45g of glucose? C 6 H 12 O 6  2C 2 H 5 OH + 2CO 2 (6x12)+12+(6x16)  2((2x12) ) 2( g of glucose would make 23g of ethanol ________ The balanced equation Data RAM C 12 O 16 H 1 Data for calculating RMM’s Simple proportion- 90  ( 46 ) 92 Calculate the formula masses of glucose and ethanol

1:Reacting masses - Na 2 CO 3 Sodium carbonate is manufactured in a two-stage process, via sodium hydrogen carbonate, as shown below. NaCl + NH 3 + CO 2 + H 2 O  NaHCO 3 + NH 4 Cl 2NaHCO 3  Na 2 CO 3 + H 2 O + CO 2 Calculate the maximum mass of sodium carbonate which could be obtained from 800 g of sodium chloride. Sodium carbonate is manufactured in a two-stage process, via sodium hydrogen carbonate, as shown below. NaCl + NH 3 + CO 2 + H 2 O  NaHCO 3 + NH 4 Cl 2NaHCO 3  Na 2 CO 3 + H 2 O + CO 2 Calculate the maximum mass of sodium carbonate which could be obtained from 800 g of sodium chloride. Sodium carbonate is manufactured in a two-stage process, via sodium hydrogen carbonate, as shown below. NaCl + NH 3 + CO 2 + H 2 O  NaHCO 3 + NH 4 Cl 2NaHCO 3  Na 2 CO 3 + H 2 O + CO 2 Calculate the maximum mass of sodium carbonate which could be obtained from 800 g of sodium chloride.

NaCl + NH 3 + CO 2 + H 2 O  NaHCO 3 + NH 4 Cl 2NaHCO 3  Na 2 CO 3 + H 2 O + CO 2 Example 1:Reacting masses - Na 2 CO 3 Tackle the 2 stages separately - less likely to make mistakes Stage 1NaCl : NaHCO 3 = NaCl + NH 3 + CO 2 + H 2 O  NaHCO 3 + NH 4 Cl 2NaHCO 3  Na 2 CO 3 + H 2 O + CO 2 Moles NaCl = mass M r Moles NaHCO 3 =  Moles NaCl = Stage 1NaCl : NaHCO 3 = 1:1 Moles NaCl = mass =800 M r 58.5 Moles NaCl = mass =800 = mol M r 58.5 Moles NaHCO 3 = 1  Moles NaCl =Moles NaHCO 3 = 1  Moles NaCl = mol

Moles Na 2 CO 3 = Mole Ratio  Moles NaHCO 3 = Stage 2 NaHCO 3 : Na 2 CO 3 = Mass Na 2 CO 3 = moles  M r = = Now use the number of moles of NaHCO 3 to calculate the moles and mass of Na 2 CO 3 NaCl + NH 3 + CO 2 + H 2 O  NaHCO 3 + NH 4 Cl 2NaHCO 3  Na 2 CO 3 + H 2 O + CO 2 NaCl + NH 3 + CO 2 + H 2 O  NaHCO 3 + NH 4 Cl 2NaHCO 3  Na 2 CO 3 + H 2 O + CO 2 Mass Na 2 CO 3 = moles  M r = 6.84  106 = Mass Na 2 CO 3 = moles  M r = 6.84  106 = 725 g Example 1:Reacting masses - Na 2 CO 3 MORELESS Stage 2 NaHCO 3 : Na 2 CO 3 = 2:1 Moles Na 2 CO 3 = 1/2 or 2/1  Moles NaHCO 3 = Moles Na 2 CO 3 = 1/2  Moles NaHCO 3 = Moles Na 2 CO 3 = 1/2  Moles NaHCO 3 = ½  Moles Na 2 CO 3 = 1/2  Moles NaHCO 3 = ½  = 6.84 mol

Empirical Formula

The empirical formula is the simplest ratio of the different atoms in it. For example for ethane it is CH 3. You can work out the empirical formula of a compound if you know the mass of each element in it. This is the reverse of finding out the % mass of a compound from its formula.

Example: What is the empirical formula of a compound that contains 27.3% carbon and 72.7% oxygen by mass? Step 1 Find, and write down the relative atomic masses of the elements involved. Ar (C) = 12 and Ar (O) = 16 Step 2 Assume that you have 100g of the compound (this makes the maths easy).  In 100g there is 27.3g of carbon and 72.7g of oxygen

Step 3 Work out how many moles of each element this must be using number of moles =Ar__ mass number of moles of carbon= 27.3 = number of moles of oxygen = 72.7=

Step 4 Divide both numbers of moles by the smallest number (2.275 in this case). number of moles of carbon = 2.275= = number of moles of oxygen = 4.55 =  In the empirical formula, there are 1 carbon atom and 2 oxygen atoms, so it is CO2.

Note: You can use the same method if you are told how many grams of each element react together to make the compound. In this case, you might be told that 27.3g of carbon react with 72.7g of oxygen atoms. You start at step 3. Another way of asking the question is to tell you how many grams of each element there are in the compound. Again, start from step 3.

Now have a go at the worksheet…