# Moles, mass and gas calculations

##### Intros

###### Lessons

##### Examples

###### Lessons

**Calculate the masses and volumes of reactants and products used in chemical reactions.**

Consider the reaction:2C$_6$H$_{14\;(l)} +$19O$_{2\;(g)} \enspace$→$\enspace$12CO$_{2\;(g)} +$14H$_2$O$_{\;(g)}$ - If 75g of C$_6$H$_{14}$ is burned, what mass of CO$_2$ would get produced from this reaction?
- If 240g of H$_2$O is produced from this reaction, how many moles of C$_6$H$_{14}$ would be required?
- At STP, what volume of O$_2$ would be required to produce 85 L of CO$_2$ in this process?
- What mass of C$_6$H$_{14}$ would be required to produce 15 moles of CO$_2$?

**Calculate the masses and volumes of reactants and products used in chemical reactions.**

Consider the reaction:P$_{4\;(s)} +$5O$_{2\;(g)} \enspace$→$\enspace$P$_4$O$_{10\;(s)}$ **Calculate the volume and number of moles of reactants involved in chemical reactions.**

Consider the reaction:2NH$_{3\;(aq)} \,+ \,$NaOCl$_{\;(aq)} \enspace$→$\enspace$N$_2$H$_{4\;(aq)} \,+ \,$NaCl$_{\;(aq)} \,+ \,$H$_2$O$_{\;(l)}$ **Calculate the volume of reactants required in a chemical process.**

Consider the reaction:SiCl$_{4\;(g)} \, + \,$2H$_{2\;(g)} \enspace$→$\enspace$Si$_{\;(s)} \, + \,$4HCl$_{\;(g)}$

500 mg of Si is required from this process. What is the total volume, in L, of H$_2$ and SiCl$_4$ required to produce this?

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###### 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.

__Notes:__- 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*10^{23}atoms) of that pure substance. - For example, one mole of
^{12}C 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. - For example, see the reaction:
- The formula: $n(mol) =$ $\large \frac{mass(g)}{M_R (\frac{g}{mol})}$ 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 M_{R}of that substance. This applies to elements and molecules, for which you can just add up the elemental mass (e.g. NH_{3}= 14+1+1+1=17).

With the moles formula above, you can remember it using the units too, like you would in general algebra: $\large\frac{g}{\frac{g}{mol}}$ = $mol$ and the g will cancel out. You are then left with $\large\frac{1}{\frac{1}{mol}}$ =$\, mol$ 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). - 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.
- 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: - Worked example: Using Avogadro’s law
At STP, 10 L of O
_{2}gas contains 0.45 mol of O_{2}molecules. If 3.5 moles more O_{2}gas is added to the container, what is the new volume?

We use the

__number of moles to relate amounts of substance in reactions__.

_{4 (g)}+ 2O

_{2 (g)}$\,$→$\,$ CO

_{2 (g)}+ 2H

_{2}O

_{ (g) }

We know that in this reaction, one mole of CH

_{4}reacts with two moles of O

_{2}. CH

_{4}has a molar mass of 16 grams per mol (g mol

^{-1}) and O

_{2}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 O

_{2}will be needed to react with 17 grams of CH

_{4}.

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.

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!

_{2 (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:

_{2}

Each fraction in the calculation above is a conversion factor: 24.3 grams of Mg metal is “the 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!

This is an observation of ideal gas behaviour. When we say “ideal 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.

We are being asked for volume (V

_{2}) so we will be solving for it using the expression above, rearranged for V

_{2}:

Using this we can apply V

_{1}= 10L, n

_{1}= 0.45 mol and n

_{2}= 3.95 mol, because 3.5 mol has been added to the 0.45 mol already in the container. This makes the expression:

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