Reacting quantities and equations

(Mass-mass problems)

PreviousPage: Calculating Molecular Masses

Our final step now, is to combine the mass scale with our earlier equation work so that we can calculate the mass of reactants and/or products taking part in a reaction.

Given substances A + B → C + D, if I want to makex grams (or kilograms, or tonnes, or…) of D, how much A and B should I startwith? In order to solve these problems you have to find the common groundbetween A and D and between B and D. (Assume in this case that C is water orsome other side product we really don’t care about.) This brings us smack upagainst moles. One mole of A reacts to formone mole of D. The ratios aren’t always1-to-1, but they always really happen in moles, not grams, not millilitres.

We can use the reaction between methane gas (which is often produced from pig manure) and oxygen gas as an example.

The balanced equation for this reaction is:

CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)

The equation tells us the nature of the reactants, products, phases and the proportions in which the reactants and products are present, so the equation means:

one molecule of CH4(g) + two molecules O2(g) → one molecule CO2(g) + two molecules of H2O(g).

Note: If we had one molecule of CH4(g) + 20 molecules O2(g) we could still only produce one molecule of CO2(g) and two molecules of H2O(g).

Excess of one reactant will not change the amount of product produced.

If we had 4 molecules of CH4(g) + 2 molecules of O2(g) → How much product would be produced?

If we had 3 molecules of CH4(g) + 6 molecules of O2(g) → How much product is produced?

If we had 6 molecules of CH4(g) + 18 molecules of O2(g) → How much product is produced?

These examples demonstrate that the coefficients in the reaction equation represent the ratio in which products react. It does not matter how many of each type of molecule is present, the reactants can only combine in that particular ratio (or multiple of that ratio).

Answers (numbers of molecules):1 & 2, 3 & 6, 6 & 12.

Examining the line of logic further:

If 1 CH4(g) molecule + 2 O2(g) molecules → 1 CO2(g) molecule + 2H2O(g) molecules.

Then 100 CH4(g) molecules + 200 O2(g) molecules → 100 CO2(g) molecules + 200 H2O(g) molecules.

Or 1x106 CH4(g) molecule + 2x106 O2(g) molecules → 1x106 CO2(g) molecule + 2x106 H2O(g) molecules

Or 1 x 6.02x1023 CH4(g) molecule + 2x6.02 x 1023 O2(g) molecules → 1 x 6.02x1023 CO2(g) molecule + 2 x 6.02x1023 H20(g) molecules.

As long as the ratio is maintained, 

any number of molecules can be used.

1 mole of CH4(g) + 2 mole of O2(g) → 1 mole of CO2(g) + 2 mole of H2O(g).

This relationship is handy because our previous work has given a method for converting moles into mass.

So 16g CH4(g) + 64g O2 (g) → 44g CO2(g) + 36g H2O(g)

But what if we do not start 1 mole of CH4(g) or 2 mole of O2(g)?

Example 1:

100g of methane burns with excess oxygen gas, what mass of CO2(g) is produced?

Step 1 Write the balanced equation CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)

Step 2 Convert mass to moles.

Step 3 Examine the equation to gain the reacting ratio.

The ratio of methane to carbon dioxide in our equation is 1:1,

ie, 1 mole CH4 → 1 mole CO2

Step 4 ∴ No. of moles of CO2 produced = 1 x 6.25 mole = 6.25 mole

Step 5 Convert moles to mass.

Mass CO2 = 6.25 mole x 44g mole-1 = 275g

Example 2:

If 100 g of CH4(g) burns in oxygen what mass of O2 is consumed?

Step 1 Write the balanced equation
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)

Step 2

Step 3 From the equation in Exercise 1 the ratio of CH4:O2 is 1:2.

Step 4 ∴ No. of moles of O2 = 2 x 6.25 mole = 12.5 mole

Step 5 Mass of O2 = 12.5mole x 32g mole-1 = 400g

Example 3:

If 50.00g of Mg(s) burns in oxygen, what mass of O2 is consumed?

Step 1 2Mg(s) + O2(g) → 2MgO(s)

Step 2

Step 3 From the equation Mg:O2 is 2:1.

So Mg:O2 is 1:½

Step 4 No. of moles of O2 = ½x2.10 moles = 1.05 moles

Step 5 Mass of O2 = 1.05 moles x 32 g mole = 33.6g.

Mass-mass problems get their name from the fact that they usually give you the mass of one substance and require the mass of another.

All mass-mass problems can be solved by following these steps:

  1. Write a balanced equation for the chemical reaction involved.
  2. Change the mass given into moles.
  3. Compare the moles of the substance given to the moles of the substance for which the mass is asked.
  4. Calculate the number of moles required.
  5. Change moles asked for into mass asked for.

  In the same way you always have to find time for problems of motion in two directions, so you must always find the number of moles to solve mass-mass problems.

Further examples.

Now try these problems.

Next Page: Concentration

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