By definition, all acids are harmful for human beings. true or false

A compound is composed of C, H and O. A 1.621 g sample of this compound was combusted, producing 1.902 g of water and 3.095 g of

vlada-n [284]

Answer: The molecular of the compound is, $C_2H_3O$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=3.095g$

Mass of $H_2O=1.902g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in $3.095g$ of carbon dioxide, $\frac{12}{44}\times 3.095=0.844g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in $1.902g$ of water, $\frac{2}{18}\times 1.092=0.121g$ of hydrogen will be contained.

For calculating the mass of oxygen:

Mass of oxygen in the compound = $(1.621)-[(0.844)+(0.121)]=0.656g$

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.844g}{12g/mole}=0.0703moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.121g}{1g/mole}=0.121moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.656g}{16g/mole}=0.041moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is $0.041$ moles.

For Carbon = $\frac{0.0703}{0.041}=1.71\approx 2$

For Hydrogen = $\frac{0.121}{0.041}=2.95\approx 3$

For Oxygen = $\frac{0.041}{0.041}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 2 : 3 : 1

Hence, the empirical formula for the given compound is $C_2H_3O_1=C_2H_3O$

The empirical formula weight = 2(12) + 3(1) + 1(16) = 43 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

$n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}$

$n=\frac{46.06}{43}=1$

Molecular formula = $(C_2H_3O_1)_n=(C_2H_3O_1)_1=C_2H_3O$

Therefore, the molecular of the compound is, $C_2H_3O$

6 0

1 year ago

Compound A is an organic compound which contains Carbon, Hydrogen and Oxygen. When 0.240g of the vapour of A is slowly passed ov

vesna_86 [32]

(a) In this section, give your answers to three decimal places.

(i)

Calculate the mass of carbon present in 0.352 g of CO

2

.

Use this value to calculate the amount, in moles, of carbon atoms present in 0.240 g

of

A

.

(ii)

Calculate the mass of hydrogen present in 0.144 g of H

2

O.

Use this value to calculate the amount, in moles, of hydrogen atoms present in 0.240 g

of

A

.

(iii)

Use your answers to calculate the mass of oxygen present in 0.240 g of

A

Use this value to calculate the amount, in moles, of oxygen atoms present in 0.240 g

of

A

(b)

Use your answers to

(a)

to calculate the empirical formula of

A

thank you

hope it helpsss

4 0

2 years ago

A mixture of Na2CO3 and MgCO3 of mass 7.63 g is reacted with an excess of hydrochloric acid. The CO2 gas generated occupies a vo

Svetlanka [38]

Answer:

58.6 % by mass of Na₂CO₃

Explanation:

This is the reaction:

Na₂CO₃ + MgCO₃ + 4HCl → MgCl₂ + 2NaCl + 2CO₂ + 2H₂O

Let's find out the moles of CO₂ produced, by the Ideal Gases Law

1.24 atm . 1.67 L = n . 0.082 . 299K

(1.24 atm . 1.67 L / 0.082 . 299K) = n

0.0844 moles = n

Ratio is 2:1, so 2 moles of dioxide were produced by 1 mol of sodium carbonate. Let's make a rule of three:

2 moles of CO₂ were produced by 1 mol of Na₂CO₃

Then, 0.0844 moles of Co₂ would beeen produced by (0.0844 .1)/2 = 0.0422 moles of Na₂CO₃.

Let's convert this moles into mass (mol . molar mass)

0.0422 mol . 106 g/mol = 4.47 g

Finally we can know the mass percent of sodium carbonate in the mixture

(Mass of compound /Total mass) . 100 → (4.47 g / 7.63g) . 100 = 58.6 %

3 0

1 year ago

Which diastereomer of oct−4−ene yields a single meso compound, (4r, 5s)−4,5−dibromooctane on reaction with br2?

stepladder [879]

How does that mean that the number is a little too late and you don’t have a phone number so you please look at it

4 0

1 year ago

Use the following half-reactions to design a voltaic cell: Sn4+(aq) + 2 e− → Sn2+(aq) Eo = 0.15 V Ag+(aq) + e−→ Ag(s) Eo = 0.80

AVprozaik [17]

Answer:

E° = 0.65 V

Explanation:

Let's consider the following reductions and their respective standard reduction potentials.

Sn⁴⁺(aq) + 2 e⁻ → Sn²⁺(aq) E°red = 0.15 V

Ag⁺(aq) + e⁻ → Ag(s) E°red = 0.80 V

The reaction with the highest reduction potential will occur as a reduction while the other will occur as an oxidation. The corresponding half-reactions are:

Reduction (cathode): Ag⁺(aq) + e⁻ → Ag(s) E°red = 0.80 V

Oxidation (anode): Sn²⁺(aq) → Sn⁴⁺(aq) + 2 e⁻ E°red = 0.15 V

The overall cell potential (E°) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.

E° = E°red, cat - E°red, an = 0.80 V - 0.15 V = 0.65 V

4 0

2 years ago