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Semmy [17]
3 years ago
13

Upon combustion, a compound containing only carbon and hydrogen produces 2.67 g

"TexFormula1" title="CO_{2}" alt="CO_{2}" align="absmiddle" class="latex-formula"> and 1.10 gH_{2}O. Find the empirical formula of the compound.
Chemistry
1 answer:
Hoochie [10]3 years ago
8 0

Answer:

\boxed{\text{CH$_{2}$}}

Explanation:

1. Calculate the mass of each element

\text{Mass of C} = \text{2.67 g } \text{CO}_{2}\times \dfrac{\text{12.01 g C}}{\text{44.01 g }\text{CO}_{2}}= \text{0.7286 g C}\\\\\text{Mass of H} = \text{1.10 g }\text{H$_{2}$O}\times \dfrac{\text{2.016 g H}}{\text{18.02 g } \text{{H$_{2}$O}}} = \text{0.1231 g H}

2. Calculate the moles of each element

\text{Moles of C = 0.7286 g C}\times\dfrac{\text{1 mol C}}{\text{12.01 g C }} = \text{0.060 67 mol C}\\\\\text{Moles of H = 0.1231 g H} \times \dfrac{\text{1 mol H}}{\text{1.008 g H}} = \text{0.1221 mol H}

3. Calculate the molar ratios

Divide all moles by the smallest number of moles.

\text{C: } \dfrac{0.06067}{0.06067}= 1\\\\\text{H: } \dfrac{0.1221}{0.06067} = 2.015

4. Round the ratios to the nearest integer

C:H = 1:2

5. Write the empirical formula

The empirical formula is \boxed{\text{CH$_{2}$}}

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Answer:

A = -213.09°C

B = 15014.85 °C

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Explanation:

Given data:

Initial volume of gas = 5.00 L

Initial temperature = 0°C  (273 K)

Final volume = 1100 mL, 280 L, 87.5 mL

Final temperature = ?

Solution:

Formula:

The given problem will be solve through the Charles Law.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

Conversion of mL into L.

Final volume = 1100 mL/1000 = 1.1 L

Final volume =  87.5 mL/1000 = 0.0875 L

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

T₂ = V₂T₁ / V₁

T₂ = 1.1 L × 273 K / 5.00 L

T₂ = 300.3 L.K / 5.00 K

T₂ = 60.06 K

60.06 K - 273 = -213.09°C

2)

V₁/T₁ = V₂/T₂

T₂ = V₂T₁ / V₁

T₂ = 280 L × 273 K / 5.00 L

T₂ = 76440 L.K / 5.00 K

T₂ = 15288 K

15288 K - 273 = 15014.85 °C

3)

V₁/T₁ = V₂/T₂

T₂ = V₂T₁ / V₁

T₂ = 0.0875 L × 273 K / 5.00 L

T₂ = 23.8875 L.K / 5.00 K

T₂ = 4.78 K

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