question 2
2.1 define the term molar mass. (2)
2.2 calculate the percentage of carbon in cane sugar c₁₂h₂₂o₁₁. (2)
2.3 cₓhᵧ is a compound made up of carbon (c) and hydrogen (h) only. 1,97g of this compound reacts with excess o₂ (g) to produce 6,2g of co₂ (g). determine the empirical formula of this compound. (6)
2.4 a metal m forms two chlorides, mcl₂ and mclₓ, that contain 55,9% and 65,5% chlorine by mass respectively. determine the:
2.4.1 relative atomic mass of m (4)
2.4.2 value of x. show all working. (4)
18

Question

Accepted Answer

### 2.1
# Brief Explanations:
Molar mass is the mass of one mole of a substance (atom, molecule, ion, etc.), expressed in grams per mole (g/mol). It is numerically equal to the substance's relative atomic or molecular mass (sum of relative atomic masses of constituent atoms) in atomic mass units (amu), scaled to grams per mole. For example, molar mass of $ \ce{O2} $ is $ 2 \times 16.00 = 32.00 \, \text{g/mol} $ (since atomic mass of O is ~16.00 amu).
# Answer:
Molar mass is the mass of one mole of a substance (in g/mol), numerically equal to its relative atomic/molecular mass (sum of atomic masses of constituent atoms) in amu, scaled to grams per mole.

### 2.2
# Explanation:
## Step1: Find molar mass of $ \ce{C12H22O11} $
Atomic masses: $ C = 12.01 \, \text{g/mol} $, $ H = 1.008 \, \text{g/mol} $, $ O = 16.00 \, \text{g/mol} $.
Molar mass $ M = (12 \times 12.01) + (22 \times 1.008) + (11 \times 16.00) $
$ M = 144.12 + 22.176 + 176.00 = 342.296 \, \text{g/mol} $

## Step2: Calculate mass of C in 1 mole
Mass of C: $ 12 \times 12.01 = 144.12 \, \text{g/mol} $

## Step3: Find % of C
$ \% C = \frac{\text{Mass of C in compound}}{\text{Molar mass of compound}} \times 100 $
$ \% C = \frac{144.12}{342.296} \times 100 \approx 42.10\% $
# Answer:
The percentage of carbon in $ \ce{C12H22O11} $ is approximately $ \boldsymbol{42.10\%} $ (or ~42.1%).

### 2.3
# Explanation:
## Step1: Find moles of $ \ce{CO2} $
Molar mass of $ \ce{CO2} = 12.01 + 2 \times 16.00 = 44.01 \, \text{g/mol} $
Moles of $ \ce{CO2} = \frac{6.2 \, \text{g}}{44.01 \, \text{g/mol}} \approx 0.1409 \, \text{mol} $

## Step2: Moles of C in compound
1 mole $ \ce{CO2} $ has 1 mole C, so moles of C in $ \ce{C_xH_y} = 0.1409 \, \text{mol} $
Mass of C = $ 0.1409 \, \text{mol} \times 12.01 \, \text{g/mol} \approx 1.692 \, \text{g} $

## Step3: Mass of H in compound
Mass of $ \ce{C_xH_y} = 1.97 \, \text{g} $, so mass of H = $ 1.97 - 1.692 = 0.278 \, \text{g} $

## Step4: Moles of H
Molar mass of H = $ 1.008 \, \text{g/mol} $
Moles of H = $ \frac{0.278 \, \text{g}}{1.008 \, \text{g/mol}} \approx 0.2758 \, \text{mol} $

## Step5: Find mole ratio (C:H)
Divide moles by smallest (C: $ 0.1409 $, H: $ 0.2758 $)
C: $ \frac{0.1409}{0.1409} = 1 $
H: $ \frac{0.2758}{0.1409} \approx 1.957 \approx 2 $

Thus, empirical formula is $ \ce{CH2} $.
# Answer:
The empirical formula of $ \ce{C_xH_y} $ is $ \boldsymbol{\ce{CH2}} $.

### 2.4.1
# Explanation:
## Step1: For $ \ce{MCl2} $ (55.9% Cl by mass)
Let relative atomic mass of M be $ A_r $.
Molar mass of $ \ce{MCl2} = A_r + 2 \times 35.45 $ (Cl = 35.45 g/mol)
Mass of Cl in $ \ce{MCl2} = 2 \times 35.45 = 70.90 \, \text{g/mol} $
% Cl = $ \frac{70.90}{A_r + 70.90} \times 100 = 55.9 $

## Step2: Solve for $ A_r $
$ \frac{70.90}{A_r + 70.90} = \frac{55.9}{100} $
$ 70.90 \times 100 = 55.9(A_r + 70.90) $
$ 7090 = 55.9A_r + 55.9 \times 70.90 $
$ 7090 = 55.9A_r + 3963.31 $
$ 55.9A_r = 7090 - 3963.31 = 3126.69 $
$ A_r = \frac{3126.69}{55.9} \approx 55.9 $ (or ~56, consistent with Fe, but we calculate as is)

(Note: For $ \ce{MCl_x} $ with 65.5% Cl, we can verify, but 2.4.1 uses $ \ce{MCl2} $.)
# Answer:
The relative atomic mass of M is approximately $ \boldsymbol{55.9} $ (or ~56, e.g., Fe).

### 2.4.2
# Explanation:
## Step1: For $ \ce{MCl_x} $ (65.5% Cl by mass)
Molar mass of $ \ce{MCl_x} = 55.9 + x \times 35.45 $ (from 2.4.1, $ A_r \approx 55.9 $)
Mass of Cl in $ \ce{MCl_x} = 35.45x \, \text{g/mol} $
% Cl = $ \frac{35.45x}{55.9 + 35.45x} \times 100 = 65.5 $

## Step2: Solve for $ x $
$ \frac{35.45x}{55.9 + 35.45x} = \frac{65.5}{100} $
$ 35.45x \times 100 = 65.5(55.9 + 35.45x) $
$ 3545x = 65.5 \times 55.9 + 65.5 \times 35.45x $
$ 3545x = 3661.45 + 2322.975x $
$ 3545x - 2322.975x = 3661.45 $
$ 1222.025x = 3661.45 $
$ x = \frac{3661.45}{1222.025} \approx 3 $

(Check: $ \ce{MCl3} $, molar mass = 55.9 + 3×35.45 = 55.9 + 106.35 = 162.25. % Cl = $ \frac{106.35}{162.25} \times 100 \approx 65.5\% $, which matches.)
# Answer:
The value of $ x $ is $ \boldsymbol{3} $.

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QUESTION IMAGE

Question

Explanation:

2.1

Step1: Find molar mass of \( \ce{C12H22O11} \)

Step2: Calculate mass of C in 1 mole

Step3: Find % of C

Step1: Find moles of \( \ce{CO2} \)

Step2: Moles of C in compound

Step3: Mass of H in compound

Step4: Moles of H

Step5: Find mole ratio (C:H)

Answer:

2.2