All categories

4 years ago

A sample of a compound contains 160 g of oxygen and 20.2 g of hydrogen. Give the compound's empirical formula.

Chemistry

2 answers:

Stella [2.4K]4 years ago

4 0

Answer : The empirical formula of a compound is, $H_2O_1$

Solution : Given,

Mass of O = 160 g

Mass of H = 20.2 g

Molar mass of O = 16 g/mole

Molar mass of H = 1 g/mole

Step 1 : convert the given masses into moles.

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{160g}{16g/mole}=10moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{20.2g}{1g/mole}=20.2moles$

Step 2 : For the mole ratio, divide each value of the moles by the smallest number of moles calculated.

For O = $\frac{10}{10}=1$

For H = $\frac{20.2}{10}=2.02\approx 2$

The ratio of O : H = 1 : 2

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula = $H_2O_1$

Therefore, the empirical formula of a compound is, $H_2O_1$

Maslowich4 years ago

3 0

Amount of oxygen in the compound = 160 g
Amount of oxygen in the compound = 20.2 gm
Mole of oxygen in the compound = 160/16
= 10 moles
Mole of hydrogen in the compound = 20.2/1.01
= 20 moles
Then
The ratio of oxygen to ration of hydrogen = 1:2
So
The empirical formula of the compound is H2O. I hope the answer has come to your help.

You might be interested in

Which example is a weak base? <br> ethyl alcohol, ammonia, potassium hydroxide or vinegar?

BlackZzzverrR [31]

Answer:

potassium

Explanation:

b/c electronegativitu

7 0

3 years ago

Carbon dioxide dissolves in water to form carbonic acid, which is primarily dissolved CO2. Dissolved CO2 satisfies the equilibri

lorasvet [3.4K]

Explanation:

The reaction equation will be as follows.

$CO_{2}(aq) + H_{2}O \rightleftharpoons H^{+}(aq) + HCO^{-}_{3}(aq)$

Calculate the amount of $CO_{2}$ dissolved as follows.

$CO_{2}(aq) = K_{CO_{2}} \times P_{CO_{2}}$

It is given that $K_{CO_{2}}$ = 0.032 M/atm and $P_{CO_{2}}$ = $1.9 \times 10^{-4}$ atm.

Hence, $[CO_{2}]$ will be calculated as follows.

$[CO_{2}]$ = $K_{CO_{2}} \times P_{CO_{2}}$

= $0.032 M/atm \times 1.9 \times 10^{-4}atm$

= $0.0608 \times 10^{-4}$

or, = $0.608 \times 10^{-5}$

It is given that $K_{a} = 4.46 \times 10^{-7}$

As, $K_{a} = \frac{[H^{+}]^{2}}{[CO_{2}]}$

$4.46 \times 10^{-7} = \frac{[H^{+}]^{2}}{0.608 \times 10^{-5}}$

$[H^{+}]^{2}$ = $2.71 \times 10^{-12}$

$[H^{+}]$ = $1.64 \times 10^{-6}$

Since, we know that pH = $-log [H^{+}]$

So, pH = $-log (1.64 \times 10^{-6})$

= 5.7

Therefore, we can conclude that pH of water in equilibrium with the atmosphere is 5.7.

3 0

3 years ago

An amount of nitrogen undergoes a physical change and increases its volume. What can be known about this physical change?

Stella [2.4K]

The answer for The question is (D)The density changed

4 0

4 years ago

Read 2 more answers

[25 pts] When a 14.2g sample of a compound containing only mercury and oxygen is decomposed, 13.2g of Hg is obtained. What is th

Agata [3.3K]

Answer:

% composition of Hg = 93 %

% composition of O = 7 %

Explanation:

Data Given:

mass of total sample = 14.2 g

mass of mercury (Hg) = 13.2 g

percent composition of Hg =

percent composition of O =

Solution:

First find mass of oxygen

As this compound only have two components that is oxygen and mercury

upon decomposition it gives 13.2 g mercury

now to find mass of oxygen we have to subtract mass of mercury from total mass of sample

Mass of oxygen = Total mass of sample - mass of Mercury

Mass of oxygen = 14.2 g - 13.2 g

Mass of oxygen = 1 g

Now to find percent composition of Hg and Oxygen

% composition = component mass / total mass of sample x 100. . . .(1)

For % composition of mercury Hg

Put values in equation 1

% composition of Hg = 13.2 g / 14.2 g x 100

% composition of Hg = 93%

For % composition Oxygen (O)

Put values in equation 1

% composition of O = 1 g / 14.2 g x 100

% composition of O = 7 %

6 0

3 years ago

Im not sure how to do this can someone help with these?

IceJOKER [234]

Answer:

1. 280 g of CO

2. 16.4 g of O₂

3. 42 g of Cl₂

Explanation:

Ans 1

Data Given:

moles of O₂= 5 moles

mass of CO = ?

Solution:

To solve this problem we have to look at the reaction

Reaction:

2CO + O₂ -----------> 2CO₂

2 mol 1 mol

So if we look at the reaction 2 mole of CO react with 1 mole of O₂ then how many moles of CO will react with 5 moles of O₂

For this apply unity formula

2 mole of CO ≅ 1 mole of O₂

X mole of CO≅ 5 mole of O₂

By Doing cross multiplication

moles of CO = 2 moles x 5 moles / 1 mol

moles of CO = 10 mole

Now calculate mass of 10 moles of CO

Formula used

mass in grams = no. of moles x Molar mass

Molar mass of CO = 12 + 16 = 28 g/mol

Put values in above formula

mass in grams = 10 moles x 28 g/mol

mass in grams = 280 g

So,

280 g of CO will react with 5 moles of O₂

_________________________

Ans 2

Data Given:

mass of C₃H₈ = 22.4 g

moles of O₂= ?

Solution:

To solve this problem we have to look at the reaction

Reaction:

C₃H₈ + 5O₂ -----------> 3CO₂ + 4H₂O

1 mol 5 mol

Convert moles to mass

Molar mass of C₃H₈ = 3(12) + 8(1)

Molar mass of C₃H₈ = 36 + 8 = 44 g/mol

and

molar mass of O₂ = 2(16) = 32 g/mol

So,

C₃H₈ + 5O₂ -----------> 3 CO₂ + 4H₂O

1 mol (44 g/mol) 5 mol (32 g/mol)

44 g 160 g

So if we look at the reaction 44 g of C₃H₈ react with 160 g of O₂, then how many grams of O₂ will react with 22.4 g of ethane

For this apply unity formula

44 g of C₃H₈ ≅ 60 g of O₂

grams of O₂ ≅ 22.4 g of ethane

By Doing cross multiplication

grams of O₂ = 22.4 g x 44 g/ 60 g

grams of O₂ = 16.4 g

16.4 g of O₂ react with 22.4 grams of ethane

______________________

Ans 3

Data Given:

mass of Rubidium Chlorate = 10 g

moles of O₂= ?

Solution:

To solve this problem we have to look at the reaction

Reaction:

2 RbClO₃ ------------ 2 RbCl + 3O₂

2 mol 3 mol

Convert moles to mass

Molar mass of RbClO₃ = 85.5 + 35.5 + 3(16)

Molar mass of RbClO₃ = 169

and

molar mass of O₂ = 2(16) = 32 g/mol

So,

2 RbClO₃ ------------> 2 RbCl + 3O₂

2 mol ( 169 g/mol) 3 mol (32 g/mol)

338 g 96 g

So if we look at the reaction 338 g of RbClO₃ gives 96 g of O₂, then how many grams of O₂ will be given by 10 g of RbClO₃

For this apply unity formula

338 g of RbClO₃ ≅ 96 g of O₂

grams of O₂ ≅ 10 g of RbClO₃

By Doing cross multiplication

grams of O₂ = 338 g x 10 g/ 96 g

grams of O₂ = 35.2 g

35.2 g of O₂ will be produce by 10 grams of RbClO₃

______________________

Ans 4

Data Given:

mass of K = 46 g

moles of Cl₂= ?

Solution:

To solve this problem we have to look at the reaction

Reaction:

2K + Cl₂ ------------> 2KCl

2 mol 1 mol

Convert moles to mass

Molar mass of K = 39 g/mole

and

molar mass of Cl₂ = 2(35.5) = 71 g/mol

So,

2K + Cl₂ ------------> 2KCl

2 mol ( 39 g/mol) 1 mol (71 g/mol)

78 g 71 g

So if we look at the reaction 78 g of K react wit 71 g of Cl₂, then how many grams of Cl₂ will react with 46 g of K

For this apply unity formula

78 g of K ≅ 71 g of Cl₂

46 g of K ≅ X grams of Cl₂

By Doing cross multiplication

grams of Cl₂ = 71 g x 46 g/ 78 g

grams of Cl₂ = 42 g

42 g of Cl₂ will react with 46 grams of K

4 0

3 years ago