Question

Your lab partner has made the observation that you always take the mass of chemicals in lab, but then use mole ratios to balance the equation. “Why not use the masses in the equation?”, your partner asks. What if your lab partner decided to balance equations by using masses as coefficients? Is this even possible? Why or why not?

Short answer

The stated way is impossible to directly determine the mass of$N{H}_3$ that could produce by 40 g of$N_2$ . But by using mole as a coefficient, it is possible to find the mass of$N{H}_3$ produced in the reaction.

Step-by-step solution

Using masses and mole ratios to balance the equation

In a balanced chemical equation, it may be required to calculate the mass of reactants and products. The ratio of compounds present in the equation can be used as a conversion factor. For the calculation of mass-to-mass conversion, one has to use the mole as an intermediate for accuracy. It is not possible to directly calculate the mass of reactant to the mass of product; one has to first convert the mass into moles.

It is not possible to balance an equation by using masses as coefficients as mass-to-mass conversion directly is not possible and is not the correct way to calculate. One has to first balance the chemical equation on the basis of moles and then it needs to be converted into masses.

Example

The balanced equation of $N{H}_3$formation:

${\text{N}}_{\text{2}}\left(\text{g}\right){\text{+ 3H}}_{\text{2}}\left(\text{g}\right)\to {\text{2NH}}_{\text{3}}\left(\text{g}\right)$

According to mole:

Ratio$\to$(1Mole) (2mole)

If we want to calculate the mass of$N{H}_3$produced by 40g of $N_2$:

There is no direct way to compare the mass of $N_2$to the mass of $N{H}_3$, it can be calculated by moles only.

Number of moles of the given compound:

$$\begin{gathered} N_2=\frac{40g}{28.0 g/mol} \\ {N}_2=1.42 \text{mol} \end{gathered}$$

And, 1 mole of $N_2$produces 2 mole of$N{H}_3$

So, 1 mole of $N_2$produces $\left(2\times 17\right)g$moles of $N{H}_3$.

Therefore, 1.42 mole of$N_2$ produce $\left(\frac{2\times 17}{1.42}\right)g$of $N{H}_3$.

Conclusion

It was impossible to directly determine the mass of$N{H}_3$ that could be produce by 40g of $N_2$. But, by using mole as a coefficient, it is possible to find the mass of $N{H}_3$produced in the reaction.