# Question 69194

Dec 23, 2015

$\text{4.8 g}$

#### Explanation:

The trick here is to realize that since the gas is collected over water, you are actually dealing with a gaseous mixture of carbon monoxide and water vapor at a total pressure of $\text{743 mmHg}$.

In order to get the partial pressure of carbon monoxide in this mixture, you're going to have to find the vapor pressure of water at ${25}^{\circ} \text{C}$ and subtract it from the total pressure of the mixture - think Dalton's Law of partial pressures.

At ${25}^{\circ} \text{C}$, water has a vapor pressure of about $\text{23.7 mmHg}$.

http://www.endmemo.com/chem/vaporpressurewater.php

This means that the partial pressure of the carbon monoxide will be equal to

${P}_{\text{mixture}} = {P}_{C O} + {P}_{{H}_{2} O}$

${P}_{C {O}_{2}} = \text{743 mmHg" - "23.7 mmHg" = "719.3 mmHg}$

From this point on, your strategy will be to use the ideal gas law to find how many moles of carbon monoxide were produced by this decomposition reaction.

$\textcolor{b l u e}{P V = n R T}$

Plug in your values and solve for $n$, the number of moles of gas - do not forget to convert the pressure from mmHg to atm and the temperature from degrees Celsius to Kelvin!

$P V = n R T \implies n = \frac{P V}{R T}$

n = (719.3/760color(red)(cancel(color(black)("atm"))) * 2.68color(red)(cancel(color(black)("L"))))/(0.0821 (color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * (273.15 + 25)color(red)(cancel(color(black)("K")))) = "0.1036 moles CO"

Now, formic acid, $\text{HCO"_2"H}$ undergoes decomposition in the presence of acids (for example, sulfuric acid) to form carbon monoxide, $\text{CO}$, and water, $\text{H"_2"O}$, in $1 : 1$ mole ratios.

${\text{HCO"_2"H"_text((aq]) stackrel(color(white)(x)color(red)("H"_2"SO"_4)color(white)(xxx))(->) "H"_2"O"_text((l]) + "CO}}_{\textrm{\left(g\right]}} \uparrow$

This means that every mole of formic acid that undergoes decomposition will produce $1$ mole of carbon monoxide. So, if the reaction produced $0.1036$ moles of $\text{CO}$, it must have consumed $0.1036$ moles $\text{HCO"_2"H}$.

Now simply use formic acid's molar mass to determine how many grams would contain that many moles

0.1036 color(red)(cancel(color(black)("moles CHO"_2"H"))) * "46.03 g"/(1color(red)(cancel(color(black)("mole CHO"_2"H")))) = "4.769 g"

Rounded to two sig figs, the number of sig figs you have for the temperature of the gas, the answer will be

m_"formic acid" = color(green)("4.8 g")#