Answer to Question #95847 in General Chemistry for Brittany Wallace

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 ${C_2}{H_4}{O_2}$.