Do Stoichiometry

In a chemical reaction, matter can neither be created nor destroyed, so the products that come out of a reaction must equal the reactants that go into a reaction. Stoichiometry is the measure of the elements within a reaction.[1] It involves calculations that take into account the masses of reactants and products in a given chemical reaction. Stoichiometry is one half math, one half chemistry, and revolves around the one simple principle above - the principle that matter is never lost or gained during a reaction. The first step in solving any chemistry problem is to balance the equation.

Steps

Balancing the Chemical Equation

  1. Write down the number of atoms that comprise each compound on either side of the equation. Using the chemical equation you can identify the atoms of each element in the reaction. Because a chemical reaction can never create or destroy new matter, a given equation is unbalanced if the number (and types) of atoms on each side of the equation don't perfectly match.
    • Don’t forget to multiply through by a coefficient or subscript if one is present.
    • For example, H2SO4 + Fe ---> Fe2(SO4)3 + H2
    • On the reactant (left) side of the equation there are 2 H, 1 S, 4 O, and 1 Fe.
    • On the product (right) side of the equation there are 2H, 3 S, 12 O, and 2 Fe.
  2. Add a coefficient in front of elements that are not oxygen and hydrogen to balance each side. Identify the lowest common factor between all elements that aren’t oxygen and hydrogen to get equal numbers of atoms on both sides.
    • For example, the lowest common factor between 2 and 1 is 2 for Fe. Add a 2 in front of the Fe on the left side to balance it.
    • The lowest common factor between 3 and 1 is 3 for S. Add a 3 in front of H2SO4 to balance the left and right sides.
    • At this stage, our equation looks like this: 3 H2SO4 + 2 Fe ---> Fe2(SO4)3 + H2
  3. Balance the hydrogen and oxygen atoms. Hydrogen and oxygen atoms are balanced last because they commonly appear in multiple molecules on each side of the equation. At this stage in balancing the equation, don’t forget to recount the atoms if you’ve added coefficients to molecules.
    • In our example, we added a 3 in front of H2SO4 and now have 6 hydrogens on the left and only 2 on the right side of the equation. We also have 12 oxygen on the left and 12 oxygen on the right, so it is balanced.
    • We can balance hydrogens by adding a 3 in front of H2.
    • Our final balanced equation is 3 H2SO4 + 2 Fe ---> Fe2(SO4)3 + 3 H2.
  4. Re-count the number of atoms on each side of the equation to make sure they are equal. Once you've finished, it's smart to go back and check the equation for balance. This can be done by again adding up all of the atoms on either side of the equation to make sure they are equal on both sides.
    • Let's check our equation, 3 H2SO4 + 2 Fe ---> Fe2(SO4)3 + 3 H2, for balance.
    • On the left side of the arrow, there are 6 H, 3 S, 12 O, and 2 Fe.
    • On the right side of the arrow, there are 2 Fe, 3 S, 12 O, and 6 H.
    • The left and the right sides of the equation match, therefore, it is now balanced.

Converting Between Grams and Moles

  1. Calculate the molar mass of the compound given in grams. The molar mass is the amount in grams (g) of one mole of a compound.[2] It allows you to easily convert between grams and moles of a substance.[3] To calculate the molar mass you need to identify how many molecules of an element are in the compound and the atomic mass of each element in the compound.
    • Define the number of atoms of each element in a compound. For example, glucose is C6H12O6, there are 6 carbon atoms, 12 hydrogen atoms, and 6 oxygen atoms.
    • Identify the atomic mass in grams per mol (g/mol) of each atom. The atomic masses of the elements in glucose are: carbon, 12.0107 g/mol; hydrogen, 1.007 g/mol; and oxygen, 15.9994 g/mol.
    • Multiply each elements atomic mass by the number of atoms present in the compound. Carbon: 12.0107 x 6 = 72.0642 g/mol; Hydrogen: 1.007 x 12 = 12.084 g/mol; Oxygen: 15.9994 x 6 = 95.9964 g/mol.
    • Adding these products together yields the molar mass of the compound. 72.0642 + 12.084 + 95.9964 = 180.1446 g/mol. 180.14 grams is the mass of one mole of glucose.
  2. Convert grams of a substance to moles using molar mass. Using the molar mass as a conversion factor, you can calculate the number of moles present in the stated number of grams of the species. Divide the known amount of grams (g) by the molar mass (g/mol).[4],[5] An easy way to check that you’ve done the correct math is to make sure the units cancel out leaving only moles.
    • For example: How many moles are in 8.2 grams of hydrogen chloride (HCl)?
    • The atomic mass of H is 1.0007 and Cl is 35.453 making the molar mass of the compound 1.007 + 35.453 = 36.46 g/mol.
    • Dividing the number of grams of the substance by the molar mass yields: 8.2 g / (36.46 g/mol) = 0.225 moles of HCl.
  3. Determine the molar ratio between reactants. In order to determine the yield of a product in a given reaction you need to determine the molar ratio. The molar ratio tells you the ratio in which substances react with each other and is given by the coefficient of the species in the balanced1 reaction.[6]
    • For example, what is the molar ratio of KClO3 to O2 in the reaction 2 KClO3 ---> 2 KCl + 3 O2.
    • First, check to see the equation is balanced. Never forget this step or your ratios will be wrong. In this case there are equal amounts of each element on both sides of the reaction so it is balanced.
    • The ratio of KClO3 to O2 is 2/3. It doesn’t matter which number goes on top or on bottom as long as you keep the same compounds on the top and bottom throughout the rest of the problem.[7]
  4. Cross-multiply by the molar ratio to find moles of other reactant. To calculate the number of moles of a species produced or required for a reaction, you will use the molar ratio.[7] Problems will usually ask you to determine the number of moles required or the number of moles produced in a reaction given a certain number of grams of reactant.
    • For example, given the reaction N2 + 3 H2 ---> 2 NH3 how many moles of NH3 will be produced given 3.00 grams of N2 reacting with sufficient H2?
    • In this example, sufficient H2 means that there is enough available and you don’t have to take it into account to solve the problem.
    • First, convert grams of N2 to moles. The atomic mass of nitrogen is 14.0067 g/mol so the molar mass of N2 is 28.0134 g/mol. Dividing mass by molar mass gives you 3.00 g/28.0134 g/mol = 0.107 mol.
    • Set up the ratios given by the question: NH3: N2 = x/0.107 mol.
    • Cross multiply this ratio by the molar ratio of NH3 to N2: 2:1. x/0.107 mol = 2/1 = (2 x 0.107) = 1x = 0.214 mol.
  5. Convert moles back to mass using the molar mass of the species. You will use molar mass again, but this time you will multiply to convert moles back to grams. Be sure to use the molar mass of the correct species.
    • The molar mass of NH3 is 17.028 g/mol. Therefore 0.214 mol x (17.028 grams/mol) = 3.647 grams of NH3.

Converting Between Liters of Gas and Moles

  1. Determine if the reaction is happening at standard temperature and pressure (STP). STP is the set of conditions given where 1 mole of an ideal gas will occupy 22.414 liters (L) of volume.[8] Standard temperature is 273.15 Kelvin (K) and standard pressure is 1 atmosphere (atm).[9]
    • Generally, a reaction will say that it is given at 1 atm and 273 K or will simply say STP.
  2. Use the conversion factor 22.414 L/mol to convert liters of gas to moles. If your reaction is happening at STP, you can use 22.414 L/mol to calculate the number of moles in a given volume of gas.[10] Divide the volume of gas (L) by the conversion factor to determine moles.
    • For example, convert 3.2 liters of N2 gas to moles: 3.2 L/22.414 L/mol = 0.143 moles.
  3. Use the ideal gas law to convert liters of gas not at STP. If you are given a reaction that does not occur at STP, you will have to use the ideal gas law PV = nRT to determine the number of moles in the reaction.[8] P is pressure in atmospheres, V is volume in liters, n is the number of moles, R is the gas law constant 0.0821 L-atm/mol-degree, and T is the temperature in degrees Kelvin.[11]
    • The equation can be rearranged to solve for moles: n = RT/PV.
    • The units of the gas constant are designed to cancel out the units of the other variables.
    • For example, determine the number of moles in 2.4 liters of O2 at 300 K and 1.5 atm. Plugging in the variables yields: n = (0.0821 x 300)/(1.5 x 2) = 24.63/3.6 = 6.842 moles of O2

Converting Between Liters of Liquid and Moles

  1. Calculate the density of the liquid. Sometimes chemical equations will give you the volume of a liquid reactant and require the amount of grams or moles needed for the reaction. To convert to grams, you will use the density of that liquid. Density is given by mass/volume.[12]
    • If the density is not given within the problem, you may have to look it up in a reference text or online.
  2. Convert the volume to milliliters (mL). To convert the volume of liquid to mass (g) you will need to the use the density of that liquid. Density is given in grams per milliliter (g/mL), therefore, the volume of the liquid needs to be in milliliters to convert.
    • Identify the volume given. For example, let’s say the problem states that you have 1 liter of H2O. To convert to mL simply multiply by 1000. There are 1000 milliliters in a liter of water.
  3. Multiply the volume by the density. When you multiply a volume (mL) by the density of that liquid (g/mL), the milliliters cancel out and you are left with grams of the substance.[11]
    • For example, the density of H2O is 18.0134 g/mL. If the chemical equation states that you have 500 mL of H2O, then the amount of grams is 500 mL x 18.0134 g/mL is 9006.7 g.
  4. Calculate the molar mass of the reactant. The molar mass is the amount in grams (g) of one mole of a compound. It allows you to easily convert between grams and moles of a substance. To calculate the molar mass you need to identify how many molecules of an element are in a compound and the atomic mass of each element in the compound.
    • Define the number of atoms of each element in a compound. For example, glucose is C6H12O6, there are 6 carbon atoms, 12 hydrogen atoms, and 6 oxygen atoms.
    • Identify the atomic mass in grams per mol (g/mol) of each atom. The atomic masses of the elements in glucose are: carbon, 12.0107 g/mol; hydrogen, 1.007 g/mol; and oxygen, 15.9994 g/mol.
    • Multiply each elements atomic mass by the number of atoms present in the compound. Carbon: 12.0107 x 6 = 72.0642 g/mol; Hydrogen: 1.007 x 12 = 12.084 g/mol; Oxygen: 15.9994 x 6 = 95.9964 g/mol.
    • Adding these products together yields the molar mass of the compound. 72.0642 + 12.084 + 95.9964 = 180.1446 g/mol. 180.14 grams is the mass of one mole of glucose.
  5. Convert grams of a substance to moles using molar mass. Using the molar mass as a conversion factor, you can calculate the number of moles present in the stated number of grams of the species. Divide the known amount of grams (g) by the molar mass (g/mol).[4],[5] An easy way to check that you’ve done the correct math is to make sure the units cancel out leaving only moles.
    • For example: How many moles are in 8.2 grams of hydrogen chloride (HCl)?
    • The atomic mass of H is 1.0007 and Cl is 35.453 making the molar mass of the compound 1.007 + 35.453 = 36.46 g/mol.
    • Dividing the number of grams of the substance by the molar mass yields: 8.2 g / (36.46 g/mol) = 0.225 moles of HCl.

Sources and Citations

  1. ↑ http://www.chemteam.info/Stoichiometry/What-is-Stoichiometry.html
  2. ↑ http://preparatorychemistry.com/Bishop_molar_mass_conversion_factors_help.htm
  3. ↑ http://www.chemteam.info/Stoichiometry/Mole-Mass.html
  4. ↑ 4.0 4.1 http://www.chemteam.info/Stoichiometry/Mass-Mass.html
  5. ↑ 5.0 5.1 https://www.khanacademy.org/science/chemistry/chemical-reactions-stoichiome/stoichiometry-ideal/a/stoichiometry
  6. ↑ http://www.chemteam.info/Stoichiometry/Molar-Ratio.html
  7. ↑ 7.0 7.1 http://www.chemteam.info/Stoichiometry/Mole-Mole.html
  8. ↑ 8.0 8.1 http://chemistry-reference.com/stoichiometry/
  9. ↑ http://www.engineeringtoolbox.com/stp-standard-ntp-normal-air-d_772.html
  10. ↑ http://www.thegeoexchange.org/chemistry/stoichiometry/liters-to-moles.html
  11. ↑ 11.0 11.1 http://www.chemtutor.com/mols.htm
  12. ↑ http://uncw.edu/chem/Courses/Reeves/OnLineLabs/scienceMajors/Density_PH.pdf

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