A mixture of anhydrous sodium carbonate and sodium hydrogencarbonate of mass 10.000g was heated until it reached a constant mass of 8.708g. Calculate the composition of the mixture in grams of each component. Sodium hydrogencarbonate thermally decomposes to form sodium carbonate.
First of all, let's convert these names of compounds to their chemical formulas:
$$ \begin{gather} \text{Anhydrous Sodium Carbonate: } \ce{Na2CO3} \\ \text{Sodium Hydrogen Carbonate: } \ce{NaHCO3} \end{gather} $$
Now, let's look at the other information given to us. We are told that the \(\ce{NaHCO3}\) decomposes to form \(\ce{Na2CO3}\).
Let's write a balanced equation for this:
$$\ce{2NaHCO3 -> Na2CO3 + H2O + CO2}$$
Now let's talk about something important with decomposition reactions: The \(\ce{H2O} + \ce{CO2}\) are both gases. Let's rewrite this formula:
$$ \ce{2NaHCO3 -> Na2CO3 + H2O_{(g)} + CO2_{(g)}} $$
This explains why the system has lost mass! Let's calculate how much mass we have lost:
$$ 10g - 8.708g = 1.292g $$
This means, the \(\ce{H2O} + \ce{CO2}\) are of mass 1.292g.
You may think we can now say that the \(\ce{Na2CO3}\) has mass 8.708g but we can't, because we don't know what percentage of the original mixture was anhydrous sodium carbonate (one of the reactants).
What we can do though, is imagine the two gaseous products as "one compound". Let's say that the \(\ce{H2O} + \ce{CO2}\) are in one system of \(\ce{H2OCO2}\). Essentially we are working out the moles of gas.
Calculate the moles of \(\ce{H2OCO2}\):
$$ \begin{gather} \text{moles} = \frac{\text{mass}}{\text{M}_\text{r}} \\\\ \text{moles} = \frac{1.292g}{18+12+(16 \times 2)} \\ \text{moles} \approx 0.020677 mol \end{gather} $$
Calculate the molar ratio of \(\ce{H2OCO2}\):\(\ce{NaHCO3}\). From the formula, we can see it is 1:2.
$$ \begin{gather} \therefore \text{mol}(\ce{NaHCO3}) = 2 \times \text{mol}(\ce{H2CO2}) \\ \text{mol}(\ce{NaHCO3}) \approx 0.041355 \end{gather} $$
We can now calculate the mass of \(\ce{NaHCO3}\).
$$ \begin{gather} \text{mol} = \frac{\text{mass}}{\text{M}_\text{r}} \\ \text{mol} \times \text{M}_\text{r} = \text{mass} \\ \text{M}_\text{r}(\ce{NaHCO3}) = 23+1+12+(16 \times 3) \\ \text{mass} \approx \text{M}_\text{r}(\ce{NaHCO3}) \times 0.041355 \\ \text{mass} \approx 84 \times 0.041355 \\ \text{mass} \approx 3.5g \\ \end{gather} $$
A mixture of calcium carbonate and magnesium carbonate with a mass of 10.000g was heated to constant mass, with final mass being 5.096g.
Firstly, we will form the two decomposition reactions:
$$ \begin{gather} \ce{MgCO3 -> MgO + CO2} \\ \ce{CaCO3 -> CaO + CO2} \end{gather} $$
The next step, is to realise that we have no idea how many moles of each carbonate we have! The formulas we just wrote assume we have 1 mol of each. That means we should form a new set of equations using \(x\) and \(y\):
$$ \begin{gather} x\ce{MgCO3 ->} x\ce{MgO} + x\ce{CO2} \\ y\ce{CaCO3 ->} y\ce{CaO} + y\ce{CO2} \end{gather} $$
Now let's join all of these together to make one huge formula, keeping this algebraic terms:
$$ x\ce{MgCO3} + y\ce{CaCO3 ->} x\ce{MgO} + y\ce{CaO} + x\ce{CO2} + y\ce{CO2} $$
Now let's go back to the question from before ad note down what the masses of these are!
We are told the mass of Calcium Carbonate and Magnesium Carbonate
sums to 10.000g.
We are told the mass of the end product is 5.096g. Because \(CO2\)
is a gas, we know that it has left the system. \(\therefore\) the
Oxides sum to give 5.096g.
Leaving 4.904g for the CO2.
Using this, we can form a set of simultaneous equations:
$$ \begin{gather} x\ce{CO2} + y\ce{CO2} = 4.904 \\ y\ce{MgO} + y\ce{CaO} = 5.096 \end{gather} $$
We can then substitute the \(M_r\) of these compounds to form mathematical equations:
$$ \begin{gather} (12+32)x + (12+32)y = 4.904 \\ (24.3+16)x + (40.1+16)y = 5.096 \end{gather} $$
Solving on a calculator gives:
$$ \begin{gather} x \approx 0.0732 \\ y \approx 0.0382 \end{gather} $$
Now that we have the moles of \(\ce{MgCO3}\) (\(x\)) , we can calculate the mass of it, by first calculating the \(M_r\)
$$ \begin{gather} \text{M}_\text{r}(MgCO3) = 24.3+12+16+16+16 = 84.3 \\ 84.3 \times x = 6.17076 \\ \frac{6.17076}{10} \times 100 \approx 61.7\% \end{gather} $$