a) Water - $H_2O$.

A water molecule has 2 hydrogen atoms ($H$) and 1 oxygen atom ($O$) so the molecular weight is

$2 \times \var{h} + \var{o} = \var{2 * h + o}$.

b) Methane - $CH_4$.

A methane molecule has 1 carbon atom ($C$) and 4 hydrogen atoms ($H$) so the molecular weight is

$\var{c} + 4 \times \var{h} = \var{c + 4 * h}$.

c) Glucose - $C_6H_{12}O_6$.

A glucose molecule has 6 carbon atoms ($C$), 12 hydrogen atoms ($H$) and 6 oxygen atoms ($O$) so the molecular weight is 

$6 \times \var{c} + 12 \times \var{h} + 6 \times \var{o} = \var{6 * c + 12 * h + 6 * o}$.

d) Calcium phospate - $Ca_3(PO_4)_2$.

A calcium phosphate molecule has 3 calcium atoms ($Ca$) and 2 phosphate ions ($PO_4$). We don't need to worry too much about what an ion is here, we just need to know that $(PO_4)_2$ means there are 2 lots of $PO_4$ which means a total of 2 phosporus atoms ($P$) and $4 \times 2 = 8$ oxygen atoms ($O$). Therefore, the molecular weight is 

$3 \times \var{ca} + 2 \times \var{p} + 8 \times \var{o} = \var{3 * ca + 2 * p + 8 * o}$.

To answer these questions we use the formula

$\text{molecular weight} \times \text{number of moles} = \text{mass of substance (in grams)}$.

a)

Glucose has a molecular weight of $\var{glucose}$. How much does  $\var{0.25 * a}$ mole(s) of glucose weigh in grams?

Solution:

Putting our numbers into the formula we find that $\var{0.25 * a}$ mole(s) of glucose weighs

$\var{glucose} \times \var{0.25 * a} = \var{glucose * 0.25 * a}$ grams.

b)

Sodium chloride has a molecular weight of $\var{NaCl}$. How much does $\var{0.25 * b}$ mole(s) of sodium chloride weigh in grams?

Solution:

Putting our numbers into the formula we find that $\var{0.25 * b}$ mole(s) of sodium chloride weighs

$\var{NaCl} \times \var{0.25 * b} = \var{NaCl * 0.25 * b}$ grams.

To answer these questions we use the formula

$\dfrac{\text{mass of substance (in grams)}}{\text{molecular weight}} = \text{number of moles}$.

a)

Glucose has a molecular weight of $\var{glucose}$. How many moles of glucose are there in $\var{5 * a}$ grams?

Solution:

Putting our numbers into the formula we find that there are 

$\begin{align} \dfrac{\var{5 * a}}{\var{glucose}} &= \var{(5 * a) / glucose} \text{ moles}\\ &= \var{precround(((5 * a) / glucose), 2)} \text{ moles to 2 d.p.} \end{align}$ 

in $\var{5 * a}$ grams of glucose.

b)

Sodium chloride has a molecular weight of $\var{NaCl}$. How many moles of sodium chloride are there in $\var{5 * b}$ grams?

Solution:

Putting our numbers into the formula we find that there are

$\begin{align}\dfrac{\var{5 * b}}{\var{NaCl}} &= \var{(5 * b) / NaCl} \text{ moles} \\ & = \var{precround(((5 * b) / NaCl), 2)} \text{ moles to 2 d.p.} \end{align}$.

in $\var{5 * b}$ grams of sodium chloride.

To answer these questions we use the formula

$\dfrac{\text{number of moles of substance}}{\text{volume of liquid (in litres)}} = \text{concentration (in mol/L)}$.

a) 

$\var{a}$ mole(s) of a substance are dissolved in $\var{b}$ litre(s) of a liquid to make a solution. What is the concentration of the solution in M (mol/L)?

Solution:

Putting our numbers into the formula, we find that the concentration is

$\begin{align}\dfrac{\var{a}}{\var{b}} & = \var{a / b} M \\ & = \var{precround((a / b), 2)} M \text{ to 2 d.p.} \end{align}$

b)

$\var{0.25 * c}$ mole(s) of a substance are dissolved in $\var{250 * d}$ml of a liquid to make a solution. What is the concentration of the solution in M (mol/L)?

Solution:

The formula uses the volume of liquid in litres so we first have to convert $\var{250 * d}$ml to a volume in litres. There are $1000$ml in 1L so $\var{250 * d}$ml is equal to 

$\dfrac{\var{250 * d}}{1000} = \var{250 * d / 1000}L$.

Putting our numbers into the formula, we find that the concentration is

$\begin{align}\dfrac{\var{0.25 * c}}{\var{250 * d / 1000}} & = \var{(0.25 * c) / (250 * d / 1000)} M \\ & = \var{precround(((0.25 * c) / (250 * d / 1000)), 2)} M \text{ to 2 d.p.} \end{align}$

To answer these questions, we use the formula

$\text{volume of liquid (in litres)} \times \text{concentration (in mol/L)} = \text{number of moles of substance}$.

a)

How many moles of glucose are there in $\var{a}$L of a $\var{0.25 * b}$M (mol/L) solution?

Solution:

Putting our numbers into the formula, we find that there are

$\var{a} \times \var{0.25 * b} = \var{a * 0.25 * b}$ moles

of glucose in $\var{a}$L of a $\var{0.25 * b}$M (mol/L) solution.

b)

How many moles of glucose are there in $\var{25 * c}$mL of a $\var{0.25 * d}$M (mol/L) solution?

Solution:

Our formula uses the volume of liquid in litres so first we have to convert $\var{25 * c}$mL to a volume in litres. There are 1000ml in 1L so $\var{25 * c}$mL is equal to 

$\dfrac{\var{25 * c}}{1000} = \var{25 * c / 1000}$L.

Putting our numbers into the formula, we find that there are

$\begin{align}\var{25 * c / 1000} \times \var{0.25 * d} & = \var{(25 * c / 1000) * 0.25 * d} \text{ moles} \\ & = \var{precround(((25 * c / 1000) * 0.25 * d),2 )} \text{ moles to 2 d.p.}\end{align}$

of glucose in $\var{25 * c}$mL of a $\var{0.25 * d}$M (mol/L) solution.

We can break this question up into parts. First we need to know how many moles of glucose we need to make a $\var{0.5 * b}$M solution using $\var{100 * a}$ml of water. The calculate this we use the formula

$\text{volume of liquid (in litres)} \times \text{concentration (in mol/L)} = \text{number of moles of substance}$.

This formula uses the voume in litres so we have to convert $\var{100 * a}$ml to a volume in litres. There are 1000ml in 1L so $\var{100 * a}$ml is equal to 

$\dfrac{\var{100 * a}}{1000} = \var{100 * a / 1000}$L.

Putting our numbers into the formula, we find that we need 

$\var{100 * a / 1000} \times \var{0.5 * b} = \var{(100 * a / 1000) * 0.5 * b}$ moles

of glucose. Finally, to work out the mass of glucose we need, we use the formula

$\text{molecular weight} \times \text{number of moles} = \text{mass of substance (in grams)}$.

We are told that glucose has a molecular wight of $\var{glucose}$ and we have calculated that we need $\var{(100 * a / 1000) * 0.5 * b}$ moles of glucose. Putting these numbers into the formula, we find that we need

$\begin{align}\var{glucose} \times \var{(100 * a / 1000) * 0.5 * b} & = \var{glucose * ((100 * a / 1000) * 0.5 * b)} \text{ grams} \\ & = \var{precround((glucose * ((100 * a / 1000) * 0.5 * b)), 2)} \text{ grams to 2 d.p.}\end{align}$

of glucose to make a $\var{0.5 * b}$M solution using $\var{100 * a}$ml of water.

$\var{50 * a}$ml of a $\var{0.25 * b}$M solution is diluted by adding another $\var{50 * c}$ml of liquid, what is the new concentration?

Solution:

We started with $\var{50 * a}$ml of liquid and added another $\var{50 * c}$ml so the new volume of liquid is 

$\var{50 * a} + \var{50 * c} = \var{50 * (a + c)} \text{ml.}$

$\var{50 * a}$ml is equal to

$\dfrac{\var{50 * a}}{1000} = \var{50 * a / 1000}\text{L}$

so the number of moles we started with is

$\var{50 * a / 1000} \times \var{0.25 * b} = \var{(50 * a / 1000) * 0.25 * b} \text{ moles}.$

We haven't added any more of the substance so we still have $\var{(50 * a / 1000) * 0.25 * b}$ moles but this is now dissolved in $\var{50 * (a + c)}$ml of liquid. $\var{50 * (a + c)}$ml is equal to $\var{50 * (a + c) / 1000}$L so the new concentration is 

$\begin{align} \dfrac{\var{(50 * a / 1000) * 0.25 * b}}{\var{50 * (a + c) / 1000}} & = \var{((50 * a / 1000) * 0.25 * b) / (50 * (a + c) / 1000)} \text{M} \\ & = \var{precround((((50 * a / 1000) * 0.25 * b) / (50 * (a + c) / 1000)), 2)} \text{M to 2 d.p.} \end{align}$

Challenge:

Can you think of a shorter way of doing this calculation? Don't worry if you can't, just go through the steps one by one.

We have been asked to do quite a lot of calculations in this example so let's break it up into several smaller parts. To work out the concentrations in g/L and M after dilution, we need to first work out the concentrations in g/L and M of the "stock" solution. Let's start by working out the concentration of the "stock" solution in g/L.

The "stock" solution contains $\var{stock_mass}$g of sodium chloride dissolved in $\var{stock_volume}$L of solution. To work out the concentration in g/L we use the formula

$\dfrac{\text{mass of substance (in grams)}}{\text{volume of liquid (in litres)}} = \text{concentration (in g/L)}.$

Putting in our numbers we find that the concentration of the "stock" solution in g/L is

$\dfrac{\var{stock_mass}}{\var{stock_volume}} = \var{stock_mass / stock_volume} \text{ g/L}.$

Next, we can work out the concentration of the stock solution in M (mol/L). The formula for this is

$\dfrac{\text{number of moles of substance}}{\text{volume of liquid (in litres)}} = \text{concentration (in mol/L)}.$

We can see that we need to work out how many moles of sodium chloride (NaCl) there are in $\var{stock_mass}$g. The formula for this is

$\dfrac{\text{mass of substance (in grams)}}{\text{molecular weight}} = \text{number of moles}.$

We can see that to do this, we are going to need to work out the molecular weight of sodium chloride (NaCl). We have been told the atomic weights of each of the atoms in sodium chloride (NaCl) so we can do this by adding together the atomic weights of all the atoms to get the molecular weight which is

$1 \times \var{Na} + 1 \times \var{Cl} = \var{NaCl} \text{ Da}.$

We can now use this to work out how many moles of sodium chloride there are in $\var{stock_mass}$g using the formula above. Putting in the numbers we find that there are

$\dfrac{\var{stock_mass}}{\var{NaCl}} = \var{stock_mass / NaCl} \text{ moles}$

of sodium chloride in $\var{stock_mass}$g. We can now work out the concentration of the "stock" solution in M (mol/L). The formula is 

$\dfrac{\text{number of moles of substance}}{\text{volume of liquid (in litres)}} = \text{concentration (in mol/L)}.$

We have $\var{stock_mass}$g of sodium chloride dissolved in $\var{stock_volume}$L of liquid and we have calculated that there are $\var{stock_mass / NaCl}$ moles of sodium chloride in $\var{stock_mass}$g. Putting these numbers into the formula we find that the concentration of the "stock" solution in M is

$\dfrac{\var{stock_mass / NaCl}}{\var{stock_volume}} = \var{stock_molarity} \text{ M}.$

Let's recap what we have calculated so far. We have worked out that the concentration of the "stock" solution in g/L is 

$\var{stock_mass_concentration} \text{ g/L}$

and in M is

$\var{stock_molarity} \text{ M}.$

Since we are diluting $\var{stock_volume_used}$ml of the "stock" solution, we need to work out how much sodium chloride there is in $\var{stock_volume_used}$ml of the solution using the concentrations we have calculated. We need to work this amount out in both grams and moles, let's start with grams. If we rearrange the formula 

$\dfrac{\text{mass of substance (in grams)}}{\text{volume of liquid (in litres)}} = \text{concentration (in g/L)}.$

we find that 

${\text{volume of liquid (in litres)}} \times \text{concentration (in g/L)} = \text{mass of substance (in grams)}.$

Now, $\var{stock_volume_used}$ml is equal to 

$\dfrac{\var{stock_volume_used}}{1000} = \var{stock_volume_used / 1000} \text{ L},$

so in $\var{stock_volume_used}$ml of "stock" solution, there are

$\var{stock_volume_used / 1000} \times \var{stock_mass_concentration} = \var{(stock_volume_used / 1000) * stock_mass_concentration} \text{ g}$

of sodium chloride. Similarly, we can work out the number of moles of sodium chloride using the formula

${\text{volume of liquid (in litres)}} \times \text{concentration (in mol/L)} = \text{number of moles of substance}.$

Putting in our numbers, we find that in $\var{stock_volume_used}$ml of "stock" solution, there are

$\var{stock_volume_used / 1000} \times \var{stock_molarity} = \var{(stock_volume_used / 1000) * stock_molarity} \text{ moles}$

of sodium chloride.

Finally, we can work out the concentration of the diluted solution. We know that $\var{stock_volume_used}$ml of "stock" solution is diluted to $\var{final_volume}$ml so the final volume of liquid is $\var{final_volume}$ml. We have worked out that in $\var{stock_volume_used}$ml of stock solution there are $\var{(stock_volume_used / 1000) * stock_mass_concentration}$g ($\var{(stock_volume_used / 1000) * stock_molarity}$ moles) of sodium chloride and we have not added any more sodium chloride so in our diluted solution we have $\var{(stock_volume_used / 1000) * stock_mass_concentration}$g ($\var{(stock_volume_used / 1000) * stock_molarity}$ moles) of sodium chloride in $\var{final_volume}$l of liquid. The formula for the concentration in g/L is

$\dfrac{\text{mass of substance (in grams)}}{\text{volume of liquid (in litres)}} = \text{concentration (in g/L)}.$

$\var{final_volume}$ml is equal to $\var{final_volume / 1000}$L and so the concentration of the diluted solution in g/L is

$\begin{align}\dfrac{\var{(stock_volume_used / 1000) * stock_mass_concentration}}{\var{final_volume / 1000}} & = \var{final_mass_concentration} \\ & = \var{precround(final_mass_concentration, 3)} \text{ g/L to 3 d.p.}\end{align}$

The formula for concentration in M (mol/L) is

$\dfrac{\text{number of moles of substance}}{\text{volume of liquid (in litres)}} = \text{concentration (in mol/L)}$

so the concentration of the diluted solution in M is

$\begin{align}\dfrac{\var{(stock_volume_used / 1000) * stock_molarity}}{\var{final_volume / 1000}} & = \var{final_molarity} \\ & = \var{precround(final_molarity, 3)} \text{ M to 3 d.p.}\end{align}$

A solution of glucose was prepared by diluting $\var{stock_volume_used}$ml of the "stock" glucose solution to $\var{final_volume}$ml. The concentration of glucose in the final solution was $\var{final_concentration}$mM. What was the concentration of glucose in the "stock" solution in M (mol/L)?

Solution:

First, we convert $\var{final_volume}$ml to a volume in litres. There are 1000ml in 1L so $\var{final_volume}$ml is equal to 

$\dfrac{\var{final_volume}}{1000} = \var{final_volume / 1000} \text{ L.}$

Next, we can calculate how many mmol of glucose there are in the final solution. The formula for this is 

$\text{volume of liquid (in litres)} \times \text{concentration (in mmol/L)} = \text{number of millimoles of substance}$.

[Note: This is slightly different to the formula on the wiki page for molar calculations. This is because we are currently working with mmol and mM as opposed to mol and M. If you prefer, you can first convert all values in mmol and mM to values in mol and M respectively and use the formulas as given on the wiki page. If you would like to read a little more about converting between units, see the wiki page on dimensions in the animal science section.]

Putting in our numbers we find that there are 

$\var{final_volume / 1000} \times \var{final_concentration} = \var{(final_volume / 1000) * final_concentration} \text{ mmol}$

of glucose in the final solution.

We also need to convert $\var{stock_volume_used}$ml to a volume in litres. $\var{stock_volume_used}$ml is equal to 

$\dfrac{\var{stock_volume_used}}{1000} = \var{stock_volume_used / 1000} \text{ L}.$

We didn't add any more glucose when diluting the "stock" solution so the "stock" solution also contained $\var{(final_volume / 1000) * final_concentration}$mmol of glucose in $\var{stock_volume_used / 1000}$L of solution. We calculate the concentration of the "stock" solution in mM (mmol/L) using the formula

$\dfrac{\text{number of millimoles of substance}}{\text{volume of liquid (in litres)}} = \text{concentration (in mmol/L)}$.

[Again, if you prefer, you can convert the amount of substance in mmol to an amount in mol before doing the calculation and use the formulas as given on the wiki page.]

Putting in our numbers, we find that theconcentration of the "stock" stock in mM is

$\dfrac{\var{(final_volume / 1000) * final_concentration}}{\var{stock_volume_used / 1000}} = \var{stock_molarity * 1000} \text{ mM}.$

Finally, we have been asked to give the answer in M so we have to convert $\var{stock_molarity * 1000}$mM to a concentration in M. There are 1000mM in 1M so the concentration of the "stock" solution in M is

$\begin{align}\dfrac{\var{stock_molarity * 1000}}{1000} & = \var{stock_molarity}\text{ M} \\ & = \var{precround(stock_molarity, 3)} \text{ M to 3 d.p.} \end{align}$