Moles, mass and gas calculations

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  1. Using mass and gas volume in stoichiometry calculations
  2. Recap stoichiometry basics
  3. Calculating moles
  4. Molar volume of gas and Avogadro's law.
  5. Worked example: using molar volume and unit conversions
  1. Calculate the masses and volumes of reactants and products used in chemical reactions.
    Consider the reaction:

    2C6_6H14β€…β€Š(l)+_{14\;(l)} +19O2β€…β€Š(g)_{2\;(g)} \enspace \enspace 12CO2β€…β€Š(g)+_{2\;(g)} + 14H2_2Oβ€…β€Š(g)_{\;(g)}

    1. If 75g of C6_6H14_{14} is burned, what mass of CO2_2 would get produced from this reaction?
    2. If 240g of H2_2O is produced from this reaction, how many moles of C6_6H14_{14} would be required?
    3. At STP, what volume of O2_2 would be required to produce 85 L of CO2_2 in this process?
    4. What mass of C6_6H14_{14} would be required to produce 15 moles of CO2_2?
  2. Calculate the masses and volumes of reactants and products used in chemical reactions.
    Consider the reaction:

    P4β€…β€Š(s)+_{4\;(s)} + 5O2β€…β€Š(g)_{2\;(g)} \enspace \enspace P4_4O10β€…β€Š(s)_{10\;(s)}

    1. At STP, what volume of O2_2 gas is needed to completely combust 2kg of P4_4?
    2. If 25 L of O2_2 were available, how much mass of P4_4 could be reacted with?
    3. What mass of P4_4O10_{10} would be produced by this?
  3. Calculate the volume and number of moles of reactants involved in chemical reactions.
    Consider the reaction:

    2NH3β€…β€Š(aq) + _{3\;(aq)} \,+ \, NaOClβ€…β€Š(aq)_{\;(aq)} \enspace \enspace N2_2H4β€…β€Š(aq) + _{4\;(aq)} \,+ \, NaClβ€…β€Š(aq) + _{\;(aq)} \,+ \, H2_2Oβ€…β€Š(l)_{\;(l)}

    1. If 5000 kg of hydrazine (N2_2H4_{4}) is required from this industrial process, how much ammonia gas (in L completely dissolved in solution at STP) is required as starting material?
    2. How many moles of hydrazine could be produced if only 5 kg of NaOCl was available?
  4. Calculate the volume of reactants required in a chemical process.
    Consider the reaction:

    SiCl4β€…β€Š(g) + _{4\;(g)} \, + \, 2H2β€…β€Š(g)_{2\;(g)} \enspace \enspace Siβ€…β€Š(s) + _{\;(s)} \, + \, 4HClβ€…β€Š(g)_{\;(g)}

    500 mg of Si is required from this process. What is the total volume, in L, of H2_2 and SiCl4_4 required to produce this?
    Topic Notes
    In this lesson, we will learn:
    • To recognize the format of stoichiometry test questions and calculations.
    • To recall the molar volume of gas at standard temperature and pressure and its meaning.
    • Methods to calculate number of moles of chemicals in reactions using mass, moles and volume of gas.
    • To apply Avogadro’s law in finding quantities of gas reactant.


    • We saw in Stoichiometry 1: Introduction that because atoms are so small, they cannot be physically counted when doing reactions in a regular laboratory. We use mass, in grams, as a measure of the amount of substance we have because this is related to moles, the actual number of atoms or molecules present by molar mass. Every substance has a molar mass, measured in grams per mole which is the mass of one mole (or 6.023*1023 atoms) of that pure substance.
      • For example, one mole of 12C carbon atoms weighs 12g. If you place pure carbon on a weighing balance and it reads as 12g, you have one mole of carbon atoms.

      We use the number of moles to relate amounts of substance in reactions.
      • For example, see the reaction:

      • CH4 (g) + 2O2 (g)  \,   \, CO2 (g) + 2H2O (g)

        We know that in this reaction, one mole of CH4 reacts with two moles of O2. CH4 has a molar mass of 16 grams per mol (g mol-1) and O2 has a molar mass of 32 grams per mole.
        With this information we can predict that 64 grams (2 moles with a mass of 32 grams per mole) of O2 will be needed to react with 17 grams of CH4.

      This is the important link between the amount of molecules/substance which we can’t directly measure, and the mass of substance which we can measure in a laboratory.

    • The formula: n(mol)=n(mol) = mass(g)MR(gmol)\large \frac{mass(g)}{M_R (\frac{g}{mol})} can be used to calculate the number of moles of a substance when you know the mass, and you can use the periodic table to find atomic or molecular mass MR of that substance. This applies to elements and molecules, for which you can just add up the elemental mass (e.g. NH3 = 14+1+1+1=17).
      With the moles formula above, you can remember it using the units too, like you would in general algebra: ggmol\large\frac{g}{\frac{g}{mol}} = mol mol and the g will cancel out. You are then left with 11mol\large\frac{1}{\frac{1}{mol}} = mol\, mol which cancels for the mol units.

    • While we can use mass to find number of moles, we need another relation for gases because it is often quite impractical to measure the mass of a gas. Gases are much easier to measure by their volume, or the amount of 3d space they occupy.
      • The molar volume of gas at STP, standard temperature and pressure (0°C or 273K, 100 kPa pressure) is 22.4 litres per mole (22.4 L/mol). In other words, one mole of atoms of a pure ideal gas at 0°C will fill 22.4 litres of space.
      • The molar volume of gas at room temperature (25°C, 298K) and pressure is 24 litres per mole (24 L/mol).

      These values for STP and room temperature and pressure are based on Avogadro’s law which is discussed later.

      Do you notice how a higher temperature makes the same amount of gas take up more space? Both temperature and pressure affect the volume of a gas, which is why we always quote molar volume with temperature and pressure. ALWAYS check the conditions of the reaction in your exam question. Don’t confuse the two!

    • Stoichiometry calculations involve unit conversions from one quantity given in the question to an unknown quantity:
      • To get to moles, use the equation and the molar ratios (the stoichiometric ratio) shown.
      • To get to volume, use the molar volume of gas constants.
      • To get to mass, use the atomic/molecular masses shown in the periodic table.
      You can use conversion factors just like in Introduction: Unit conversions in chemistry. For example, see the reaction:

      O2 (g) + 2Mg (s)   \,   \, 2MgO (s)

      If the reaction took place at standard temperature and pressure using 72.9 g of Mg, the volume of gas to completely react this amount of magnesium would be:

      72.9 g Mg=g \, Mg =1 mol Mg24.3 g Mgβˆ—1 mol O22 mol Mgβˆ—22.4 L O21 mol O2\large \frac{1 \, mol \, Mg} {24.3 \, g \, Mg} * \frac{1 \, mol \, O_{2}} {2 \, mol \, Mg} * \frac{22.4 \, L \, O_{2}} {1 \, mol \, O_{2}} == 33.6 L O2

      Each fraction in the calculation above is a conversion factor: 24.3 grams of Mg metal is β€œthe same” as 1 mole of Mg metal, we are just changing the units of the expression, not the value of it. The third fraction is the molar volume at standard temperature and pressure we quoted above. Using this, we’ve been able to work out the amount of substance needed for a reaction indirectly!

    • As mentioned above, the molar volumes of gas at RTP and STP are based on Avogadro’s law which states that at constant temperature and pressure, equal volumes of any two gases have the same number of moles.
      This can be expressed as:

    • V1n1=V2n2\large \frac{V_{1}} {n_{1}} = \frac{V_{2}} {n_{2}}

      This is an observation of ideal gas behaviour. When we say β€œideal gas” we are making some assumptions about real gases which are not perfectly accurate because real gases do deviate slightly. The assumptions are however very accurate at mild conditions like STP and room temperature. For more on this, see the lesson Ideal gas equation and Kinetic Molecular Theory.

    • Worked example: Using Avogadro’s law At STP, 10 L of O2 gas contains 0.45 mol of O2 molecules. If 3.5 moles more O2 gas is added to the container, what is the new volume?

    • We are being asked for volume (V2) so we will be solving for it using the expression above, rearranged for V2:

      V2=V_{2} = V1n1\large \frac{V_{1}} {n_{1}} β€‰βˆ—β€‰n2 \, * \, n_{2}

      Using this we can apply V1 = 10L, n1 = 0.45 mol and n2 = 3.95 mol, because 3.5 mol has been added to the 0.45 mol already in the container. This makes the expression:

      V2=V_{2} = 100.45β€‰βˆ—β€‰\large \frac{10} {0.45} \,* \, 3.95 = 87.8 L O2L \,O_{2}