11.5: Applications of the Ideal Gas Law- Molar Volume, Density and Molar Mass of a Gas, 11.6: Mixtures of Gases and Partial Pressures. For more information contact us at info@libretexts.org or check out our status page at https://status.libretexts.org. \begin{align} What we know: Pressure (1 atm),temperature (273 K), the identity of the gas (N2). Note that R = 0.08205 L atm /(K mol) will not be suitable in this case. This site explains how to find molar mass. common chemical compounds. When calculating molecular weight of a chemical compound, it tells us how many grams are in one mole of that substance. Stoichiometry is the theme of the previous block of modules, and the ideal gas law is the theme of this block of modules. In other words, 1 mole of a gas will occupy 22.4 L at STP, assuming ideal gas behavior. Molar Volume Formula. The formula of the molar volume is expressed as \(V_{m} = \frac{Molar\ mass}{Density} Where V m is the volume of the substance. }\end{align}\). &= \mathrm{0.20\: M\: AgNO_3} A mole of anything has the same number of identities as the number of atoms in exactly 12 grams of carbon-12, the most abundant isotope of carbon. If 500 mL of $$\ce{HCl}$$ gas at 300 K and 100 kPa dissolve in 100 mL of pure water, what is the concentration? \mathrm{V_{H_2S}} &= \mathrm{0.10\: mol \times 22.4\: L/mol}\\ V &=\dfrac{n RT}{P}\\ Legal. The Molar volume is directly proportional to molar mass and inversely proportional to density. This is not the same as molecular mass, which is the mass of a single molecule of well-defined isotopes. With this information we can calculate the density ($$\rho$$) of a gas using only its molar mass. Atomic mass: $$\mathrm{H = 1.0}$$; $$\mathrm{O = 16.0}$$; $$\mathrm{S = 32.0}$$. &= \mathrm{812^\circ C} Discussion This is how to calculate molar mass (average molecular weight), which is based on isotropically weighted averages. Using the chemical formula of the compound and the periodic table of elements, we can add up the atomic weights and calculate molecular weight of the substance. In chemistry, the formula weight is a quantity computed by multiplying the atomic weight (in atomic mass units) of each element in a chemical formula by the number of atoms of that element present in the formula, then adding all of these products together. For bulk stoichiometric calculations, we are usually determining molar mass, which may also be called standard atomic weight or average atomic mass. $$\mathrm{\dfrac{x\: g}{23.0\: g/mol}\times\dfrac{1\: mol\: H_2}{2\: mol\: Na}+\dfrac{(2.0 - x)\: g\: Ca}{40.1\: g\: Ca/mol}\times\dfrac{1\: mol\: H_2}{1\: mol\: Ca}=\dfrac{1.164\: L\: H_2 \times 100.0\: kPa}{8.3145\: kPa\: L\: mol^{-1}\: K^{-1}\: 300.0\: K}}$$, $$\mathrm{\dfrac{x}{46.0}+\dfrac{2}{40.1}-\dfrac{x}{40.1}= 0.0467\:all\: in\: mol}$$, $$\mathrm{40.1\, x + 2 \times 46.0 - 46.0\, x = 86.1}$$, $$\mathrm{-5.9\, x = 86.1 - 92.0 = -5.91}$$, $$\mathrm{Mass\: of\: Na = x = 1.0\: g}$$, $$\mathrm{Mass\: of\: Ca = 2.0 - x = 1.0\: g}$$, $$\mathrm{Mass\: Percentage\: of\: Na = 100\times \dfrac{1}{2.0} = 50\%}$$, $$\mathrm{Mole\: of\: Na = \dfrac{1}{23} = 0.0435\: mol}$$, $$\mathrm{Mole\: percentage = \dfrac{\dfrac{1}{23}}{\dfrac{1}{23} + \dfrac{1}{40.1}} = 0.635 = 63.5\%}$$, Compare this example with gravimetric analyses using the reaction, $$\mathrm{Ag^+_{\large{(aq)}} + \sideset{ }{_{\large{(aq)}}^{-}}{Cl} \rightarrow AgCl_{\large{(s)}}}$$. this can be further simplified if we work at STP: We can use these equations to identify an unknown gas, as shown below: A unknown gas has density of 1.78 g/L at STP.

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