Compound A contains carbon, hydrogen and oxygen. Analysis of 0.44 g of A gave 0.88 g of carbon dioxide and 0.36 g of water on completion of combustion in oxygen. (i) Determine the empirical formula of A. (ii) Given that the Molar Mass of A is 44 g/mol, determine its molecular formula. (iii) How can the presence of carbon in compound A be identified?

Chemistry
Compound A contains carbon, hydrogen and oxygen. Analysis of 0.44 g of A gave 0.88 g of carbon dioxide and 0.36 g of water on completion of combustion in oxygen. (i) Determine the empirical formula of A. (ii) Given that the Molar Mass of A is 44 g/mol, determine its molecular formula. (iii) How can the presence of carbon in compound A be identified?

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Answer

\ce{CH2O}

3(a) Empirical formula of compound A

Step 1: Assume 100 g100~\mathrm{g} of compound A.

MassofC=40.0 g,molesC=40.012=3.33 molMassofH=6.7 g,molesH=6.71=6.7 molMassofO=53.3 g,molesO=53.316=3.33 mol\begin{align*} &Mass of C = 40.0~g, &\quad moles C &= \frac{40.0}{12} = 3.33~mol\\ &Mass of H = 6.7~g, &\quad moles H &= \frac{6.7}{1} = 6.7~mol\\ &Mass of O = 53.3~g, &\quad moles O &= \frac{53.3}{16} = 3.33~mol \end{align*}

Step 2: Divide moles by smallest value (3.33 mol3.33~\mathrm{mol}).

C:3.333.33=1H:6.73.33=2.01≈2O:3.333.33=1\begin{align*} C: &\quad \frac{3.33}{3.33} = 1\\ H: &\quad \frac{6.7}{3.33} = 2.01 \approx 2\\ O: &\quad \frac{3.33}{3.33} = 1 \end{align*}

Empirical formula = CH2O\mathbf{CH_2O} \ceCH2O\boxed{\ce{CH2O}}

3(b) Molecular formula (molar mass = 60 g/mol60~\mathrm{g/mol})

Step 1: Molar mass of \ceCH2O\ce{CH2O} = 12+2(1)+16=30 g/mol12 + 2(1) + 16 = 30~\mathrm{g/mol}.

n=6030=2n = \frac{60}{30} = 2

Molecular formula = \ce(CH2O)2=C2H4O2\ce{(CH2O)_2} = \mathbf{C_2H_4O_2} \ceC2H4O2\boxed{\ce{C2H4O2}}

3(c) Name of compound A

\ceCH3COOH, aceticacid(ethanoicacid)\mathbf{\ce{CH3COOH},~acetic acid (ethanoic acid)} \ceCH3COOH\boxed{\ce{CH3COOH}}

2. True or false: Carbon always shows 4 covalent bonds

True (Carbon has 4 valence electrons and forms 4 covalent bonds to achieve octet stability.)

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3(a) Empirical formula of compound A Step 1: Assume 100~g of compound A.

Compound A contains carbon, hydrogen and oxygen. Analysis of 0.44 g of A gave 0.88 g of carbon dioxide and 0.36 g of water on completion of combustion in oxygen. (i) Determine the empirical formula of A. (ii) Given that the Molar Mass of A is 44 g/mol, determine its molecular formula. (iii) How can the presence of carbon in compound A be identified?
Chemistry

This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.

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3(a) Empirical formula of compound A Step 1: Assume 100~g of compound A. &Mass of C = 40.0~g, & moles C &= (40.0)/(12) = 3.33~mol\\ &Mass of H = 6.7~g, & moles H &= (6.7)/(1) = 6.7~mol\\ &Mass of O = 53.3~g, & moles O &= (53.3)/(16) = 3.33~mol Step 2: Divide moles by smallest value (3.33~mol). C: & (3.33)/(3.33) = 1\\ H: & (6.7)/(3.33) = 2.01 ≈ 2\\ O: & (3.33)/(3.33) = 1 Empirical formula = CH_2O CH2O 3(b) Molecular formula (molar mass = 60~g/mol) Step 1: Molar mass of CH2O = 12 + 2(1) + 16 = 30~g/mol. n = (60)/(30) = 2 Molecular formula = (CH2O)_2 = C_2H_4O_2 C2H4O2 3(c) Name of compound A CH3COOH,~acetic acid (ethanoic acid) CH3COOH 2. True or false: Carbon always shows 4 covalent bonds True (Carbon has 4 valence electrons and forms 4 covalent bonds to achieve octet stability.)