Question

1.
For the following reaction, 6.41 grams of oxygen gas are mixed with excess ammonia. The reaction yields 3.75 grams of nitrogen monoxide.

ammonia (g) + oxygen (g) nitrogen monoxide (g) + water (g)

What is the theoretical yield of nitrogen monoxide ? grams
What is the percent yield of nitrogen monoxide ? %

2. When 4.682 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 14.69 grams of CO2 and 6.015 grams of H2O were produced.

In a separate experiment, the molar mass of the compound was found to be 42.08 g/mol. Determine the empirical formula and the molecular formula of the hydrocarbon.

3. A 13.25 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 19.43 grams of CO2 and 7.954 grams of H2O are produced.

In a separate experiment, the molar mass is found to be 60.05 g/mol. Determine the empirical formula and the molecular formula of the organic compound.

Answer 1

Problem №1.

Let's write down the equation of chemical reaction (unbalanced)

$$N{H_3} + {O_2} = NO + {H_2}O$$

and balance it (first, balance hydrogen, then nitrogen and at the last oxygen)

$$4N{H_3} + 5{O_2} = 4NO + 6{H_2}O$$

Let's calculate the molar mass of molecular oxygen and nitrogen monoxide (atomic weights can be found in the periodic table)

$${M_{{O_2}}} = 2{M_O} = 2 \cdot 16[\frac{{\text{g}}}{{{\text{mol}}}}] = 32[\frac{{\text{g}}}{{{\text{mol}}}}]$$

$${M_{NO}} = {M_N} + {M_O} = 16[\frac{{\text{g}}}{{{\text{mol}}}}] + 14.01[\frac{{\text{g}}}{{{\text{mol}}}}] = 30.01[\frac{{\text{g}}}{{{\text{mol}}}}]$$

Now we can see from the equation, that for each $1$ mole of oxygen we can theoretically get $0.8$ moles of nitrogen monooxide (because $\frac{4}{5} = 0.8$), thus

$${\nu _{NO}} = \frac{4}{5}{\nu _{{O_2}}}$$

and using $\nu = \frac{m}{M}$ we get

$${m_{NO}} = \frac{4}{5}\frac{{{M_{NO}}}}{{{M_{{O_2}}}}}{m_{{O_2}}} = \frac{4}{5}\frac{{30.01[\frac{{\text{g}}}{{{\text{mol}}}}]}}{{32[\frac{{\text{g}}}{{{\text{mol}}}}]}} \cdot 6.41[{\text{g}}] \approx 4.81[{\text{g}}]$$

The percent yeld is

$$\eta = \frac{{{m_{em}}}}{{{m_{th}}}} \cdot 100[\% ] = \frac{{3.75[{\text{g}}]}}{{4.81[{\text{g}}]}} \cdot 100[\% ] \approx 77.96[\% ]$$

Answer: ${m_{NO}} \approx 4.81[{\text{g}}]$ , $\eta \approx 77.96[\% ]$

Problem №2.

Let's write down the equation of chemical reaction (unbalanced)

$$C{_x}{H_y} + {O_2} = C{O_2} + {H_2}O$$

and balance it

$${C_x}{H_y} + (x + \frac{y}{4}){O_2} = xC{O_2} + \frac{y}{2}{H_2}O$$

Let's calculate molar masses

$${M_{{C_x}{H_y}}} = x{M_C} + y{M_H} = (12.01x + 1.01y)[\frac{{\text{g}}}{{{\text{mol}}}}]$$ $${M_{C{O_2}}} = {M_C} + 2{M_O} = 44.01[\frac{{\text{g}}}{{{\text{mol}}}}]$$ $${M_{{H_2}O}} = 2{M_H} + {M_O} = 18.02[\frac{{\text{g}}}{{{\text{mol}}}}]$$

We can see from the equation of chemical reaction that for each 1 mole of ${C_x}{H_y}$ we get $x$ moles of $C{O_2}$ and $\frac{y}{2}$ moles of ${H_2}O$ . So we've got the following system

$$\begin{cases} {\nu _{C{O_2}}} = x{\nu _{{C_x}{H_y}}} \\ {\nu _{{H_2}O}} = \frac{y}{2}{\nu _{{C_x}{H_y}}} \end{cases}$$

and using $m = \nu M$ we get

$$\begin{cases} \frac{{14.69[{\text{g}}]}}{{44.01[\frac{{\text{g}}}{{{\text{mol}}}}]}} = x{\nu _{{C_x}{H_y}}} \\ \frac{{6.015[{\text{g}}]}}{{18.02[\frac{{\text{g}}}{{{\text{mol}}}}]}} = \frac{y}{2}{\nu _{{C_x}{H_y}}} \end{cases}$$

Simplify it

$$\begin{cases} 0.334[{\text{mol}}] =x{\nu _{{C_x}{H_y}}} \\ 0.334[{\text{mol}}] = \frac{y}{2}{\nu _{{C_x}{H_y}}} \end{cases}$$

we see that the left sides is approximately equal, it means that

$$x{\nu _{{C_x}{H_y}}} \approx \frac{y}{2}{\nu _{{C_x}{H_y}}}$$

thus

$$x = \frac{y}{2}$$

We find the empirical formula, it is $C{H_2}$. Now we need to find the chemical formula, that has the form ${C_x}{H_{2x}}$.

Use the known molar mass of the compound

$${M _{{C_x}{H_{2x}}}} = (12.01x + 1.01 \cdot 2 \cdot x)[\frac{{\text{g}}}{{{\text{mol}}}}] = 42.08[\frac{{\text{g}}}{{{\text{mol}}}}]$$

thus

$$x = \frac{{42.08[\frac{{\text{g}}}{{{\text{mol}}}}]}}{{14.03[\frac{{\text{g}}}{{{\text{mol}}}}]}} \approx 3$$

and the chemical formula is ${C_3}{H_6}$

Answer: empirical formula is $C{H_2}$ , chemical formula is ${C_3}{H_6}$

Problem №3.

We shall use the same way as in the problem 2, so we omitt some details

Let's write down the balanced chemical reaction

$${C_x}{H_y}{O_z} + (x + \frac{y}{4} - \frac{z}{2}){O_2} = xC{O_2} + \frac{y}{2}{H_2}O$$

Molar masses of reagents and products are

$${M_{{C_x}{H_y}{O_z}}} = (12.01x + 1.01y + 16z)[\frac{{\text{g}}}{{{\text{mol}}}}]$$ $${M_{C{O_2}}} = 44.01[\frac{{\text{g}}}{{{\text{mol}}}}]$$ $${M_{{H_2}O}} = 18.02[\frac{{\text{g}}}{{{\text{mol}}}}]$$

We've got the system the same system as in previous problem

$$\begin{cases} {\nu _{C{O_2}}} = x{\nu _{{C_x}{H_y}}} \\ {\nu _{{H_2}O}} = \frac{y}{2}{\nu _{{C_x}{H_y}}} \end{cases}$$

$$\begin{cases} \frac{{19.43[{\text{g}}]}}{{44.01[\frac{{\text{g}}}{{{\text{mol}}}}]}} = x{\nu _{{C_x}{H_y}}} \\ \frac{{7.954[{\text{g}}]}}{{18.02[\frac{{\text{g}}}{{{\text{mol}}}}]}} = \frac{y}{2}{\nu _{{C_x}{H_y}}} \end{cases}$$

$$\begin{cases} 0.441[{\text{mol}}] =x{\nu _{{C_x}{H_y}}} \\ 0.441[{\text{mol}}] = \frac{y}{2}{\nu _{{C_x}{H_y}}} \end{cases}$$

so again, $x{\nu _{{C_x}{H_y}}} \approx \frac{y}{2}{\nu _{{C_x}{H_y}}}$ and $y=2x$. Now let's use the known molar mass of the compound

$${M_{{C_x}{H_y}{O_z}}} = (12.01x + 1.01 \cdot 2x + 16z)[\frac{{\text{g}}}{{{\text{mol}}}}] = 60.05[\frac{{\text{g}}}{{{\text{mol}}}}]$$

Let's assume (without lose generality) that $z = kx$ , where $k$ is positive integer number. We get

$$(14.03 + 16k)x[\frac{{\text{g}}}{{{\text{mol}}}}] = 60.05[\frac{{\text{g}}}{{{\text{mol}}}}]$$

Easy to see that $k$ can't be more than 3 (because in this case $x < 1$ ) so we just need to choose between 3 cases $k = 1,2,3$. We can exclude cases $k = 2$ and $k = 3$ because they get us (with the one possible $x = 1$ ) molar masses $46.03[\frac{{\text{g}}}{{{\text{mol}}}}]$ and $62.03[\frac{{\text{g}}}{{{\text{mol}}}}]$ respectively, so the only variant is $k = 1$ and the empirical formula is $C{H_2}O$ . To get chemical formula we need to calculate x

$$(14.03 + 16 \cdot 1)x[\frac{{\text{g}}}{{{\text{mol}}}}] = 60.05[\frac{{\text{g}}}{{{\text{mol}}}}]$$

and we get $x \approx 2$ , so chemical formula is ${C_2}{H_4}{O_2}$.

Answer: empirical formula is $C{H_2}O$ and chemical formula is ${C_2}{H_4}{O_2}$.