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Chapter 6: Problem 95

A compound is \(85.6 \%\) carbon by mass. The rest is hydrogen. When \(10.0 \mathrm{g}\) of the compound is evaporated at \(50.0^{\circ} \mathrm{C},\) the vapor occupies \(6.30 \mathrm{L}\) at \(1.00 \mathrm{atm}\) pressure. What is the molecular formula of the compound?

Short Answer

Expert verified
The molecular formula of the compound is \(C_{3}H_{6}\).

Step by step solution

01

Finding the ratio of Carbon to Hydrogen

Consider that the mass of the compound is 100g (to simplify the calculation). Given that it's 85.6% carbon, the mass of carbon in 100g of the compound is \( 85.6g \) . Since the rest of the compound is hydrogen, the mass of hydrogen is \( 100g - 85.6g = 14.4g \). Find the moles of carbon and hydrogen using their atomic masses, \( C = 12.01g/mol \), \( H = 1.008g/mol \). Moles of C = \( 85.6 ÷ 12.01 ≈ 7.12 \) and Moles of H = \( 14.4 ÷ 1.008 ≈ 14.29 \). The ratio of \( C : H \) is therefore approximately \( 7 : 14 \), or more simply, \( 1:2 \).
02

Determining Molar Mass using gas law

The ideal gas law is \( PV = nRT \) , where \( P \) is pressure in atm, \( V \) is volume in L, \( n \) is the amount of substance in moles, \( R \) is the gas constant 0.0821 \( atm \cdot L/mol \cdot K \) (at these units), and \( T \) is the absolute temperature in Kelvin. Convert the given Celsius temperature to Kelvin by adding 273.15: \( 50.0 + 273.15 = 323.15K \). Rearrange the gas law equation to solve for \( n = PV/RT \), then substitute given values: \( n = (1.00atm * 6.30L) / (0.0821 * 323.15K) ≈ 0.238 moles \). Molar mass of the compound is \(10.0g / 0.238 mol ≈ 42 g/mol.
03

Determining the Molecular Formula

The empirical formula found in step 1 was \(CH_{2}\) which has a molar mass of \(12.01 g/mol + 2 * 1.008 g/mol = 14.03 g/mol \). The ratio of the molecular mass to the empirical formula mass is \(42g/mol / 14.03g/mol ≈ 3 \). Therefore, the molecular formula is \(C_{3}H_{6}\).

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Most popular questions from this chapter

If the Kelvin temperature of a sample of ideal gas doubles (e.g., from 200 K to 400 K), what happens to the root-mean-square speed, \(u_{\mathrm{rms}}\) ? (a) \(u_{\mathrm{rms}}\) increases by a factor of \(\sqrt{2} ;\) (b) \(u_{\mathrm{rms}}\) increases by a factor of \(2 ;(\mathrm{c}) u_{\mathrm{rms}}\) decreases by a factor of 2 (d) \(u_{\mathrm{rms}}\) increases by a factor of \(4 ;\) (e) \(u_{\mathrm{rms}}\) decreases by a factor of 4.

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A 10.0 g sample of a gas has a volume of \(5.25 \mathrm{L}\) at \(25^{\circ} \mathrm{C}\) and \(762 \mathrm{mm} \mathrm{Hg} .\) If \(2.5 \mathrm{g}\) of the same gas is added to this constant 5.25 L volume and the temperature raised to \(62^{\circ} \mathrm{C},\) what is the new gas pressure?

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