# Calculating mass and volume of two different substances required for one solution

I want to produce a solution, which contains $0.02~\mathrm{mol~L}^{-1}~\mathrm{Cu}^{2+}$ and $~0.2~\mathrm{mol~L}^{-1}~\mathrm{NH}_3$. The following substances are given:

• copper(II) sulfate pentahydrate
• $100~\mathrm{mL~H}_2\mathrm{O}$
• diluted ammonia solution with a concentration of $2~\mathrm{mol~L}^{-1}$.

I need to determine the required volume ($V$) of $\mathrm{NH}_3$ and the mass ($m$) of $\mathrm{CuSO}_4\cdot 5\mathrm{H}_2\mathrm{O}$.

I have no idea how to calculate $m$. However, to calculate $V$, I solved the following system of equations:

$\dfrac{2~\mathrm{mol}}{\mathrm{L}} = \dfrac{n}{V}$

$\dfrac{0.2~\mathrm{mol}}{\mathrm{L}} = \dfrac{n}{V+0.1~\mathrm{L}}$

Therefore, $V = 0.01111~\mathrm{L}$. Is this correct?

Is $m=555~\mathrm{mg}$ correct?

• To calculate the molar mass of a molecule you simply add the masses of each atom in the molecule. Eg H2O weighs 18.016 g mol^-1. So Cu is 63.5 g mol^-1, S is 32, O is 16and H is 1.008. 63.5+32+(4*16)*5*(2*1.008+16)=249.58g mol^-1 Nov 13, 2014 at 19:14
• To calculate the molar mass of a molecule you simply add the masses of each atom in the molecule. Eg H2O weighs 18.016 g mol^-1. So Cu is 63.5 g mol^-1, S is 32, O is 16and H is 1.008. 63.5+32+(4*16)*5*(2*1.008+16)=249.58g mol^-1. You want 0.02 mols L-1 or 0.002 mols in 100mls. So 0.002mol *249.58 g mol-1= 0.499g. Nov 13, 2014 at 19:20

We have direct formula for calculating the weight of the substance as, $$\text{Weight}=\frac{\text{Molarity} \times \text {molecular wt} \times \text{volume in mL}}{1000}$$ $$=\frac{0.02\times 250 \times 100}{1000}$$ $$=0.5 \text{ g}$$

So to prepare 0.02 $\mathrm{mol~L}^{-1}~\mathrm{Cu}^{2+}$ ,0.5 g of copper(II) sulfate pentahydrate should be dissolved in 100 mL water.

To find the volume of ammonia,

Initial molarity = 2 M

Initial volume= ?

Final molarity = 0.2 M

Final volume = 100 mL

We have formula, $$\text {Initial molarity} \times \text{Initial volume} = \text {Final molarity} \times \text{Final volume}$$ $$\text{Initial volume} = \frac{0.2 \times 100}{2}$$ $$= 10 mL$$

Hence, 10 mL of 2 $\mathrm{mol~L}^{-1} \mathrm{NH}_3$ should be added in 100 mL water to obtain 0.2 $\mathrm{mol~L}^{-1} \mathrm{NH}_3$.