Exercise \(\PageIndex{2.E.28}\) What is the density of laughing gas, dinitrogen monoxide, N 2 O, at a temperature of 325 K and a pressure of 113.0 kPa? Answer 1.84g /L Strategy Map Step Hint 1. Identify what information the question gives you and do any necessary conversions. See LibreText 2.4, Stoichiometry of Gaseous Substances, Mixtures, and Reactions 2. There are two approaches you could use: Approach a. To do this in one-step you can derive an equation that contains all known variables and the unknown density by manipulating the ideal gas equation to obtain an expression for density. Recall the density equation. Recall n is amount in moles which is equal to mass/molar mass. Substitute n for \(\frac{\mathrm{m}}{\mathrm{M}}\) in the ideal gas equation. Manipulate the ideal gas equation (move the variables around) so that you can solve for density. Approach b. To do this in two steps first use the ideal gas equation to solve for volume. Then use the density equation to solve for density using the calculated volume. \(\mathrm{V}=\frac{\mathrm{nRT}}{\mathrm{P}}\) \(\mathrm{d}=\frac{\mathrm{m}}{\mathrm{V}}\) 3. Plug data in to equation and solve for your density. Make sure to write your units in for all values. Solution \(113.0 \mathrm{kPa}=\frac{1 \mathrm{~atm}}{101.325 \mathrm{kPa}}=1.115 \mathrm{~atm}\) Approach a, using a derived equation: \(\mathrm{d}=\frac{\text { mass }}{\text { volume }}=\frac{\mathrm{m}}{\mathrm{v}}\) \(\begin{gathered} \mathrm{PV}=\mathrm{nRT} \\ \mathrm{n}=\frac{\text { mass }}{\text { molar mass }}=\frac{\mathrm{m}}{\mathrm{M}} \end{gathered}\) \(\begin{aligned} & \mathrm{PV}=\frac{\mathrm{m}}{\mathrm{M}} \mathrm{RT} \\ & \mathrm{d}=\frac{\mathrm{m}}{\mathrm{V}}=\frac{\mathrm{PM}}{\mathrm{RT}} \end{aligned}\) \(\begin{gathered} \mathrm{d}=\frac{(1.115 \mathrm{~atm})(44.013 \mathrm{~g} / \mathrm{mol})}{(0.08206 \mathrm{~atm} \cdot \mathrm{L} / \mathrm{mol} \cdot \mathrm{k})(325 \mathrm{k})} \\ \mathbf{d}=\mathbf{1 . 8 4 \mathrm { g } / \mathrm { L }} \end{gathered}\) Approach b, using a two steps: Step one \(\begin{aligned} &\mathrm{PV}=\mathrm{nRT}\\ &\mathrm{V}=\frac{\mathrm{nRT}}{\mathrm{P}} \end{aligned}\) \(\begin{gathered} \mathrm{V}=\frac{(1.000 \mathrm{~mol})(0.08206 \mathrm{~atm} \cdot \mathrm{L} / \mathrm{mol} \cdot \mathrm{k})(325 \mathrm{k})}{1.115 \mathrm{~atm}} \\ \mathrm{~V}=23.9 \mathrm{~L} \end{gathered}\) Step two \(\begin{gathered} d=\frac{m}{V} \\ d=\frac{44.013 g}{23.9 L} \\ d=1.84 \mathrm{~g} / L \end{gathered}\) Guided Solution Download Guided Solution as a pdf Guided Solution Hint This is a calculation type problem where you are required to calculate the density of a given compound using one of the two possible methods. See LibreText 2.4.1 (section 2.4.1 Density of a Gas) Question: What is the density of laughing gas, dinitrogen monoxide, N2O, at a temperature of 325 K and a pressure of 113.0 kPa? We are given the temperature in Kelvin, the pressure in kilopascals and the chemical formula of Dinitrogen Monoxide. From this information, we are trying to calculate the density of the dinitrogen monoxide. You will need to convert from kilopascals to atmospheres. The conversion value is: 101.325kPa = 1atm. You will need to convert from kilopascals to atmospheres. The conversion factor is: 101.325kPa = 1atm. Recall how to determine density from mass and volume. \(d=\frac{m}{V}\) How to manipulate the equation to derive a new one: 1. Recall the density equation. 2. Recall n is amount in grams per mole which is mass/molar mass. 3. Substitute (\frac{\mathrm{m}}{\mathrm{M}}\) in the ideal gas equation in place of n. 4. Manipulate the ideal gas equation (move the variables around) so that it is equal to density. We know the equation for density is \(d=\frac{m}{V}\) We know the ideal gas equation is \(\mathrm{PV}=\mathrm{nRT} \\\) We know that \(\mathrm{n}=\frac{\text { mass }}{\text { molar mass }}=\frac{\mathrm{m}}{\mathrm{M}}\) Therefore, we can substitute \(\frac{\mathrm{m}}{\mathrm{M}}\) in the ideal gas equation in place of n. \(\mathrm{PV}=\frac{\mathrm{m}}{\mathrm{M}} \mathrm{RT}\) Now, we can manipulate the equation to be equal to \(\frac{\mathrm{m}}{\mathrm{V}}\) which is equal to density (d). \(\mathbf{d}=\frac{\mathbf{m}}{\mathbf{V}}=\frac{\mathbf{P M}}{\mathbf{R T}}\) \(d=\frac{m}{V}\) NaN OR use the equations you already know: Using the ideal gas equation, plug in the information your given to solve for volume. THEN using the density equation, plug in your calculated volume. What about the mass? Recall that you can find the mass by finding the molar mass of the compound. Look at the chemical formula. Complete Solution: \(113.0 \mathrm{kPa}=\frac{1 \mathrm{~atm}}{101.325 \mathrm{kPa}}=1.115 \mathrm{~atm}\) Approach a, using a derived equation: We know the equation for density is \(\mathrm{d}=\frac{\text { mass }}{\text { volume }}=\frac{\mathrm{m}}{\mathrm{v}}\) We know the ideal gas equation is \(\mathrm{PV}=\mathrm{nRT} \\\) Now, we can manipulate the equation to be equal to (\frac{\mathrm{m}}{\mathrm{V}}\) which is equal to density (d). \(\mathbf{d}=\frac{\mathbf{m}}{\mathbf{V}}=\frac{\mathbf{P M}}{\mathbf{R T}}\) Plugging into the equation: \(\mathbf{d}=\frac{\mathbf{P M}}{\mathbf{R T}}\) You need the molar mass of N2O: \(M=2(14.007)+15.999=44.013 \mathrm{~g} / \mathrm{mol}\) \(\begin{gathered} \mathrm{d}=\frac{(1.115 \mathrm{~atm})(44.013 \mathrm{~g} / \mathrm{mol})}{(0.08206 \mathrm{~atm} \cdot \mathrm{L} / \mathrm{mol} \cdot \mathrm{k})(325 \mathrm{k})} \\ \boldsymbol{d}=\mathbf{1 . 8 4} \mathbf{g} / \mathbf{L} \end{gathered}\) Approach b, using a two steps: 1. Calculate the volume. Plug known information into the ideal gas equation; rearranged to solve for volume. \(\begin{aligned} &\mathrm{PV}=\mathrm{nRT}\\ &\mathrm{V}=\frac{\mathrm{nRT}}{\mathrm{P}} \end{aligned}\) \(\begin{gathered} \mathrm{V}=\frac{(1.000 \mathrm{~mol})(0.08206 \mathrm{~atm} \cdot \mathrm{L} / \mathrm{mol} \cdot \mathrm{k})(325 \mathrm{k})}{1.115 \mathrm{~atm}} \\ \mathrm{~V}=23.9 \mathrm{~L} \end{gathered}\) 2. Calculate the density. \(d=\frac{m}{V}\) Using the volume you calculated in the previous step, and the molar mass of N2O plug into the density equation: \(\begin{aligned} & d=\frac{44.013 g}{23.9 L} \\ & \mathbf{d}=\mathbf{1 . 8 4 g} / \mathbf{L} \end{aligned} \) Answer 1.84 g/L How to find M (molar mass) of N2O: Look up element mases on periodic table, then add up making sure you use molecule stoichiometry. N= 14.007g/mol x2 O=15.999g/mol Molar mass = 2(14.007)+15.999 = 44.013g/mol Check your work! Make sure your units cancel when you plug them in to your equation, and that your final result is in the appropriate units for density. Why does this answer make chemical sense? Density is the amount of matter that is packed into a given volume (mass divided by volume). It is a physical property that varies between materials. Gas density is the amount of matter in each volume at a specific temperature and pressure, therefore it makes sense that we can calculate the density of dinitrogen monoxide gas at these specific conditions. It does not matter which stage you include the conditions in your calculation if they are used to achieve your final solution. Using a dimensional analysis approach and writing units beside values in equations lets you see if the unis cancel to give you a result in appropriate units. Our gas density should have units of g/L. (question from page titled 9.E Gases (Exercises), shared under a CC BY 4.0 license platform https://chem.libretexts.org/Bookshelves/General_Chemistry/Chemistry_1e_(OpenSTAX)/09%3A_Gases/9.E%3A_Gases_(Exercises) , original source https://openstax.org/books/chemistry-2e/pages/9-exercises , access for free at https://openstax.org/books/chemistry-2e/pages/1-introduction )