# Using quantiative analysis to find empirical and molecular formula

A compound of Ca, C, N and S was subjected to quantitative analysis and formula mass determination, and the following data were recorded.

A $\pu{0.250 g}$ sample was mixed with $\ce{Na2CO3}$ to convert all of the Ca to $\pu{0.160 g}$ of $\ce{CaCO3}$.

A $\pu{0.115 g}$ of the compound was carried through a series of reactions until all of its S was changed to $\pu{0.344 g}$ of $\ce{BaSO4}$.

A $\pu{0.712 g}$ sample was processed to liberate all of its N as $\ce{NH3}$ and $\pu{0.155 g}$ of $\ce{NH3}$ was obtained.

The formula mass was found to be $\pu{156 g mol-1}$. Determine the empirical formula of this compound.

My issue here is that we don't really know what the starting compound is and particularly how much of each Ca, C, N, and S is there. Without knowing that, how can you balance your reaction equation and find the number of moles needed to get the empirical formula?

• Normalize all the samples to say 100 grams and work from there. The empirical formula would contain integer values of all four elements. An educated guess could be made given the four elements involved. – MaxW Oct 11 '17 at 4:09

The problem you gave doesn't quite work out, so I searched it up; did you perhaps mean $0.115 \space \text{g}$ of the sample were used in finding the sulfur content (you said $0.15 \space \text{g}$ instead).

If you did mean $0.115 \space \text{g}$, then we can figure out that the molecular formula of the unknown compound is $\text{CaC}_2\text{N}_2\text{S}_2$. Let me explain:

First, let's try figuring out how many moles of $\text{Ca}$ there are for each mole of the unknown compound, using the information that $0.250\space \text{g}$ of the compound was used to form $0.16 \space \text{g}$ of CaCO3. Converting everything to moles, this means we used $(0.250\space \text{g})\space/\space (156 \space \text{g/mol})=0.00160 \space \text{mol}$ of the unknown compound to get $(0.16\space \text{g})\space/\space (100.0869 \space \text{g/mol})=0.00160\space \text{mol CaCO}_3$, or $0.00160\space \text{mol Ca}$. This means that one mole of the unknown compound will contain $(0.00160\space \text{mol Ca})\space/\space (0.00160 \space \text{mol Unknown Compound}) \space\times\space (1 \space\text{mol Unknown Compound})=1\space\text{mol Ca}$

We can do this for each of the other procedures to get that for each mole of the unknown substance, we have $1\space\text{mol Ca}$, $2\space\text{mol S}$, and $2\space\text{mol N}$. Now, we can just subtract the masses of $\text{Ca}$, $\text{S}$, and $\text{N}$ from the molar mass of the unknown compound to get the mass of $\text{C}$ in the compound. For each mole of the unknown compound, we get that we have $\text{156 g}-\text{40.08 g}-2(\text{14.01 g})-2(\text{32.06 g})=23.8 \space\text{g}$ of carbon. This is just $(23.8\space \text{g})\space/\space (12.01 \space \text{g/mol})=1.98\space \text{mol} \approx 2 \space \text{mol}$ carbon per mole of unknown compound. So, since we have $1\space\text{mol Ca}$, $2\space\text{mol S}$, $2\space\text{mol N}$, and $2\space\text{mol C}$ for each mole of the unknown compound, the unknown compound must be $\boxed{\text{CaC}_2\text{N}_2\text{S}_2}$.

• Nicely done! OP should've indeed typed the question more accurately as with the wrong mass the formula would've been $\ce{CaS_{1.5}N2C_{3.3}}$, which doesn't make sense. – andselisk Oct 14 '17 at 17:54
• I edited original question to denote actual mass of the sample for S content, it was clearly an OP's typo. – andselisk Oct 14 '17 at 18:08

Summarizing the approach (nicely illustrated in an answer by Brandon Lee) towards solving similar quantitative analysis problems, it is always a good idea to present any reaction or set of reactions as the following conversion scheme where stoichiometric coefficient about analyzed compound is normalized to 1:

$$\ce{E^i_{x_i}R -> \frac{x_i}{a_i} E^i_{a_i}R'} \tag{1}$$

where $\ce{E^i}$ is the $i$-th element being quantitatively determined; $x_i$ and $a_i$ are stoichiometric coefficients of the given element in unknown compound and obtained product, respectively; $\ce{R}$ and $\ce{R'}$ are the rests.

For simplicity, lets abbreviate analyzed compound $\ce{E^i_{$x_i$}}R$ as $\ce{X}$ and $i$-th product $\ce{E^i_{$a_i$}R'}$ as $\ce{P_$i$}$:

$$\ce{X -> \frac{x_i}{a_i} P_i} \tag{1a}$$

According to the principle of mass conservation amounts of the $i$-th element must be preserved:

\begin{align} n_i (\ce{X}) &= \frac{x_i}{a_i} n(P_i) \tag{2} \\ \frac{m_i(\ce{X})}{M(\ce{X})} &= \frac{x_i}{a_i} \cdot \frac{m(\ce{P_$i$})}{M(\ce{P_$i$})} \tag{3} \\ x_i &= a_i \frac{m(\ce{P_$i$}) \cdot M(\ce{X})}{m_i(\ce{X}) \cdot M(\ce{P_$i$})} \tag{4} \end{align}

where $M$ is molecular weight and for the analyzed compound $\ce{X}$ is determined as

$$M(\ce{X}) = \sum_{i = 1}^n{x_i \cdot M(\ce{E^i})} \tag{5}$$

which allows to find the latter coefficient $x_n$ indirectly after all analytical routines performed for $n-1$ elements:

\begin{align} M(\ce{X}) = \sum_{i = 1}^{n-1}{x_i \cdot M(\ce{E^i})} + x_n \cdot M(\ce{E^n}) \tag{6} \\ x_n = \frac{M(\ce{X}) - \sum_{i = 1}^{n-1}{x_i \cdot M(\ce{E^i})}}{M(\ce{E^n})} \tag{7} \end{align}

For example, using the first step ($i = 1$) $\ce{E^i} = \ce{Ca}$; $\ce{R} \equiv \ce{S_{$x_2$}N_{$x_3$}C_{$x_4$}}$; $\ce{E^i_{$x_i$}R} \equiv \ce{Ca_{$x_1$}S_{$x_2$}N_{$x_3$}C_{$x_4$}}$; $\ce{R'} \equiv \ce{CO3}$; $a_1 = 1$ equation (1) can be written as

$$\ce{Ca_{x_1}S_{x_2}N_{x_3}C_{x_4} -> x_1 CaCO3}$$

and coefficient $x_1$ is therefore determined according to (4) as follows:

$$x_1 = 1 \cdot \frac{\pu{0.160 g} \cdot \pu{156 g mol-1}}{\pu{0.250 g} \cdot \pu{100 g mol-1}} = 1.0$$

Same applies to $x_2$ and $x_3$ for sulfur and nitrogen; in order to find $x_4$ (carbon content) equation (7) is used.