Ask & Answer Question

Chemistry NCERT Exemplar Solutions Class 12th Chapter Eight Chemistry Class 12th Chemistry NCERT Exemplar Solutions Class 12th Chapter One Some Basic Concepts of Chemistry

116 g of a substance upon dissociation reaction yields 7.5 g of hydrogen 60 g of oxygen and 48.5 g of carbon. Given that the atomic masses of H, O and C are 1, 16 and 12, respectively. The data agrees with how many formulae of the following?

(a) CH₃COOH (b) HCHO (c) CH₃OOCH₃ (d) CH₃CHO

3 Views | Posted 2 months ago

Asked by Shiksha User

1 Answer

Answered by

Payal Gupta | Contributor-Level 10

2 months ago

Determining empirical formula of the given compound.

$% o f C = \frac{48.5}{116} \times 100 % = 41.8 %$

$% o f H = \frac{7.5}{116} \times 100 % = 6.5 %$

$% o f O = \frac{60}{116} \times 100 % = 51.7 %$

Mass Moles Simplest ratio

C 41.8g $\frac{41.8}{12} = 3.48$ $\frac{3.48}{3.23} \approx 1$ &n

...more

Determining empirical formula of the given compound.

$% o f C = \frac{48.5}{116} \times 100 % = 41.8 %$

$% o f H = \frac{7.5}{116} \times 100 % = 6.5 %$

$% o f O = \frac{60}{116} \times 100 % = 51.7 %$

Mass Moles Simplest ratio

C 41.8g $\frac{41.8}{12} = 3.48$ $\frac{3.48}{3.23} \approx 1$

H 6.5 g $\frac{6.5}{1} = 6.5$ $\frac{6.5}{3.23} \approx 2$

O 51.7 g $\frac{51.7}{16} = 3.23$ $\frac{3.23}{3.23} = 1$

Ratio of moles of C, H and O is 1 : 2 : 1

A. CH₃COOH ⇒ C : H : O = 2 : 4 : 2 = 1 : 2 : 1

B. HCHO ⇒ C : H : O = 1 : 2 : 1

C. CH₃OOCH₃ ⇒ C : H : O = 2 : 6 : 2 = 1 : 3 : 1

D. CH₃CHO ⇒ C : H : O = 2 : 4 : 1

So, CH₃COOH and HCHO formulae agree with the given data

less

Determining empirical formula of the given compound. <math> <mrow> <mn>%</mn> <mtext> </mtext> <mtext> </mtext> <mi>o</mi> <mi>f</mi> <mtext> </mtext> <mtext> </mtext> <mi>C</mi> <mo>=</mo> <mfrac> <mrow> <mn>4</mn> <mn>8</mn> <mo>.</mo> <mn>5</mn> </mrow> <mrow> <mn>1</mn> <mn>1</mn> <mn>6</mn> </mrow> </mfrac> <mo>×</mo> <mn>1</mn> <mn>0</mn> <mn>0</mn> <mn>%</mn> <mo>=</mo> <mn>4</mn> <mn>1</mn> <mo>.</mo> <mn>8</mn> <mn>%</mn> </mrow> </math>                <math> <mrow> <mn>%</mn> <mtext> </mtext> <mtext> </mtext> <mi>o</mi> <mi>f</mi> <mtext> </mtext> <mtext> </mtext> <mi>H</mi> <mo>=</mo> <mfrac> <mrow> <mn>7</mn> <mo>.</mo> <mn>5</mn> </mrow> <mrow> <mn>1</mn> <mn>1</mn> <mn>6</mn> </mrow> </mfrac> <mo>×</mo> <mn>1</mn> <mn>0</mn> <mn>0</mn> <mn>%</mn> <mo>=</mo> <mn>6</mn> <mo>.</mo> <mn>5</mn> <mn>%</mn> </mrow> </math>                <math> <mrow> <mn>%</mn> <mtext> </mtext> <mtext> </mtext> <mi>o</mi> <mi>f</mi> <mtext> </mtext> <mtext> </mtext> <mi>O</mi> <mo>=</mo> <mfrac> <mrow> <mn>6</mn> <mn>0</mn> </mrow> <mrow> <mn>1</mn> <mn>1</mn> <mn>6</mn> </mrow> </mfrac> <mo>×</mo> <mn>1</mn> <mn>0</mn> <mn>0</mn> <mn>%</mn> <mo>=</mo> <mn>5</mn> <mn>1</mn> <mo>.</mo> <mn>7</mn> <mn>%</mn> </mrow> </math>                             Mass         Moles                  Simplest ratioC            41.8g        <math> <mrow> <mfrac> <mrow> <mn>4</mn> <mn>1</mn> <mo>.</mo> <mn>8</mn> </mrow> <mrow> <mn>1</mn> <mn>2</mn> </mrow> </mfrac> <mo>=</mo> <mn>3</mn> <mo>.</mo> <mn>4</mn> <mn>8</mn> </mrow> </math>         <math> <mrow> <mfrac> <mrow> <mn>3</mn> <mo>.</mo> <mn>4</mn> <mn>8</mn> </mrow> <mrow> <mn>3</mn> <mo>.</mo> <mn>2</mn> <mn>3</mn> </mrow> </mfrac> <mo>≈</mo> <mn>1</mn> </mrow> </math>                 H            6.5 g        <math> <mrow> <mfrac> <mrow> <mn>6</mn> <mo>.</mo> <mn>5</mn> </mrow> <mrow> <mn>1</mn> </mrow> </mfrac> <mo>=</mo> <mn>6</mn> <mo>.</mo> <mn>5</mn> </mrow> </math>             <math> <mrow> <mfrac> <mrow> <mn>6</mn> <mo>.</mo> <mn>5</mn> </mrow> <mrow> <mn>3</mn> <mo>.</mo> <mn>2</mn> <mn>3</mn> </mrow> </mfrac> <mo>≈</mo> <mn>2</mn> </mrow> </math>                      O           51.7 g        <math> <mrow> <mfrac> <mrow> <mn>5</mn> <mn>1</mn> <mo>.</mo> <mn>7</mn> </mrow> <mrow> <mn>1</mn> <mn>6</mn> </mrow> </mfrac> <mo>=</mo> <mn>3</mn> <mo>.</mo> <mn>2</mn> <mn>3</mn> </mrow> </math>         <math> <mrow> <mfrac> <mrow> <mn>3</mn> <mo>.</mo> <mn>2</mn> <mn>3</mn> </mrow> <mrow> <mn>3</mn> <mo>.</mo> <mn>2</mn> <mn>3</mn> </mrow> </mfrac> <mo>=</mo> <mn>1</mn> </mrow> </math>                  Ratio of moles of C, H and O is 1 : 2 : 1A. CH3COOH ⇒ C : H : O = 2 : 4 : 2 = 1 : 2 : 1B. HCHO ⇒ C : H : O = 1 : 2 : 1C. CH3OOCH3 ⇒ C : H : O = 2 : 6 : 2 = 1 : 3 : 1D. CH3CHO ⇒ C : H : O = 2 : 4 : 1So, CH3COOH and HCHO formulae agree with the given data

0 Upvotes 0 Downvotes

116 g of a substance upon dissociation reaction yields 7.5 g of hydrogen 60 g of oxygen and 48.5 g of carbon. Given that the atomic masses of H, O and C are 1, 16 and 12, respectively. The data agrees with how many formulae of the following?

(a) CH₃COOH (b) HCHO (c) CH₃OOCH₃ (d) CH₃CHO

1 Answer

Similar Questions for you

Learn more about...

Most viewed information

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question

116 g of a substance upon dissociation reaction yields 7.5 g of hydrogen 60 g of oxygen and 48.5 g of carbon. Given that the atomic masses of H, O and C are 1, 16 and 12, respectively. The data agrees with how many formulae of the following? (a) CH3COOH (b) HCHO (c) CH3OOCH3 (d) CH3CHO

1 Answer

Similar Questions for you

Learn more about...

Most viewed information

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question

116 g of a substance upon dissociation reaction yields 7.5 g of hydrogen 60 g of oxygen and 48.5 g of carbon. Given that the atomic masses of H, O and C are 1, 16 and 12, respectively. The data agrees with how many formulae of the following?

(a) CH₃COOH (b) HCHO (c) CH₃OOCH₃ (d) CH₃CHO