Write down the chemical equation that gives you hydrogen as a product. The problem usually tells you what the reactants and products are. For example, you may have methane and water reacting to give you carbon dioxide and hydrogen. The unbalanced chemical equation for this is CH4 + H2O gives you CO2 + H2.
Balance the chemical equation. Balancing the chemical equation involves making sure that the moles of each element are the same on both sides. For example, in the equation
CH4 + H2O gives you CO2 + H2, there are six moles of hydrogen for the reactants, but only two moles of hydrogen for the products. There is also one mole of oxygen for the reactants, but two moles of oxygen for the products. By placing the coefficient 2 before the H2O and the coefficient 4 before the H2, you have eight moles of hydrogen on both sides of the equation, and two moles of oxygen on both sides. The balanced version is
CH4 + 2H2O gives you CO2 + 4H2.
Determine the molar mass of methane (CH4), water (H2O) and hydrogen (H2). The periodic table gives the molar masses of all the elements. For CH4, the molar mass of carbon is 12.01 g / mol, and the molar mass of hydrogen is 1.008 g / mol. Since there are 4 moles of hydrogen, multiply the 1.008 g / mol by 4. Adding these together gives you,
12.01 g /mol + 4(1.008 g / mol) = 12.01 g / mol + 4.032 g / mol = 16.042 g / mol. For water, you get 2(1.008 g / mol) + 16.00 g / mol = 2.016 g / mol + 16.00 g / mol = 18.016 g / mol. For hydrogen, the molar mass is 2(1.008 g / mol) = 2.016 g / mol.
Find the amount of hydrogen produced based on the amount of methane. The problem will tell you how much of each reactant that you have. For example, you may have 60.0 grams (g) of CH4. You can use dimensional analysis for this process. For the CH4, the first ratio is 60.0 g of CH4 over 1, the second ratio is 1 mole of CH4 over 16.042 g CH4 and the third ratio is 4 moles of H2 over 1 mole of CH4. Multiplying these ratios gives you 14.961 moles of H2.
Find the amount of hydrogen produced based on the amount of water. The problem may state that you have 45.0 g of H2O. You can use dimensional analysis to find the answer. For the H2O, the first ratio is 45.0 g of H2O over 1, the second ratio is 1 mole of H2O over 18.016 g H2O and the third ratio is 4 moles of H2 over 2 moles of H2O. This gives you 4.996 moles of H2. Since the H2O gives you a smaller amount of moles of H2, as opposed to the Ch4, then the H2O is the limiting reagent.
Find the theoretical amount of hydrogen that should be produced based on the amount of hydrogen produced by the limiting reagent. For example, the 45.0 g of H2O gives you
4.996 g of H2. Multiplying this amount by 2.016 g / mol, the molar mass of H2, gives you 10.072 g of H2 produced.
Put the proper values for the amount of hydrogen into the percent yield equation. The percent yield equation will tell you how pure the hydrogen is. The formula for percent yield is: percent yield = (actual amount / theoretical amount) * 100.
The problem may tell you that 9.500 g of hydrogen was produced. The theoretical amount of hydrogen that should have been produced was 10.072 g of hydrogen. This gives you an answer of (9.500 g / 10.072 g) * 100 = 94.32 %. So, the hydrogen in this example would be 94.32% pure.