Combustion of a 0.9827 g sample of a compound containing only carbon, hydrogen, and oxygen produced 1.900 g of $CO_2$ and 1.070 g of $H_2O$. What is the empirical formula of the compound?

Md Ariful Islam

0 like 0 dislike
1 views
Share with your friends

1 Answers

Md Ariful Islam

Call

Here is the equation with masses:

$"0.9827 g C"_a"H"_b"O"_c + "(1.900 g + 1.070 g - 0.9827 g) O"_2 rarr "1.070 g H"_2"O" + "1.900 g CO"_2$

Use 32.00 g/mol for the molar mass of $"O"_2$
Use 18.03 g/mol for the molar mass of $"H"_2"O"$
Use 44.01 g/mol for the molar mass of $"CO"_2$

We do not know the molar mass of the hydrocarbon so we allow $a, b$, and $c$ to be any positive real number.

$"C"_a"H"_b"O"_c+ ("1.9873 g")/("32.00 g/mol") "O"_2 rarr ("1.070 g")/("18.03 g/mol") "H"_2"O" + ("1.900 g")/("44.01 g/mol") "CO"_2$

Perform the division:

$"C"_a"H"_b"O"_c+ ("0.06210 mol") "O"_2 rarr ("0.05935 mol") "H"_2"O" + ("0.04317 mol") "CO"_2$

Matching coefficients, we get:

$a = 0.04317$
$b = 2 xx 0.05935 = 0.1187$
$c = 0.05935 + 2 xx 0.04317 - 2 xx 0.06210 = 0.02149$

Divide every number by 0.02149:

$a/0.02149 = 2.009$
$b/0.02149 = 5.523$
$c/0.02149 = 1$

Multiply every number by 2:

$2.009 xx 2 = 4.018$
$5.523 xx 2 = 11.05$
$1 xx 2 = 2$

Round off each number to the nearest integer.

$4.018 ≈ color(white)(l)4$
$11.05 ≈ 11$
$color(white)(ll)2color(white)(ml) =color(white)(ll) 2$

The empirical formula is $"C"_4"H"_11"O"_2$.

0 like 0 dislike
1 views
Talk Doctor Online in Bissoy App
If a .500g sample of citric acid containing only carbon, hydrogen, and oxygen is burned to produce .687g $CO_2$ and .187g $H_2O$ what is the percent composition and how do you calculate it?

The idea here is that you can use the masses of carbon dioxide, $"CO"_2$, and water, $"H"_2"O"$, produced by the reaction to determine how much carbon and how much hydrogen...

1 Answers 1 views
A compound containing only carbon and hydrogen has a carbon-to-hydrogen mass ratio of 11.89. Which carbon-to-hydrogen mass ratio is possible for another compound composed only of carbon and hydrogen?

Trying to present a possible Answer We know that the molar mass of carbon and hydrogen are $C=12"g/mol" and H = 1"g/mol"$ When Carbon Hydrogen mass ratio is 11.89 The...

1 Answers 1 views
A 0.596 g sample of a gaseous compound containing only boron and hydrogen occupied 484 cm^3 at STP. When ignited in excess oxygen all the hydrogen was recovered as 1.17 g of water, and all the boron was present as B2O3. What is the empirical formula?

!! LONG ANSWER !! The trick here is to recognize the fact that you can use the mass of water produced by the reaction to determine how much hydrogen the...

1 Answers 1 views
A compound contains only $C$, $H$, and$ N$. Combustion of $35.0$ $mg$ of the compound produces $33.5$ $mg$ $CO_2$ and $41.1$ $mg$ $H_2O$. What is the empirical formula of the compound?

We can write an equation for the (complete) combustion of the substance as follows: $C_xH_yN_z + (x+y/2+z)/2O_2 to xCO_2+y/2H_2O+zNO_2$ That might look complicated, but all we're saying is that each...

1 Answers 1 views
On combustion, 1.0 L of a gaseous compound of hydrogen, carbon, and nitrogen gives 2.0 L of $CO_2$, 3.5 L of $H_2O$ vapor, and 0.50 L of $N_2$ at STP. What is the empirical formula of the compound?

The most important thing to realize here is that for gases that react under the same conditions for pressure and temperature, the is equivalent to the volume ratio. This...

1 Answers 1 views
What is the empirical formula of a compound containing $C, H$, and $O$ if combustion of 3.69 g of the compound yields 5.40 g of $CO_2$ and 2.22 g of $H_2O$?

ONLY the $C$ and the $H$ of the combustion can be presumed to derive from the unknown. (Why? Because the analysis is performed in air and typically an oxidant is...

1 Answers 1 views
A sample of a compound weighs 200 grams and contains only carbon, hydrogen, oxygen, and nitrogen. By analysis, it contains 97.56 grams of carbon, 52.03 g of oxygen and the rest is nitrogen. How do you its empirical formula?

You haven't given us the constituent percentage of hydrogen. Would you review the question? Once we have $%H$ we can address the question.

1 Answers 1 views
A compound with a molar mass of 180 g is made of carbon, hydrogen and oxygen. When it is burned, the products are 44.0 g of $CO_2$, and 8 g of $H_2O$. Based on this, what are the empirical formula and the molecular formula of the compound?

To solve this problem, we look at the amount of products. 44 g of $CO_2$ is one mole of this compound, and so, contains one mole of carbon atoms and...

1 Answers 1 views
Combustion analysis of a 12.01 g sample of tartaric acts - which contains only carbon, hydrogen, and oxygen - produced 14.08 g CO2 and 4.32 g H2O. What is the empirical formula for tartaric acid?

We address the molar quantities of carbon, and hydrogen .... convert these into masses, and then address the empirical formula. $"Moles of carbon dioxide"=(14.01*g)/(44.01*g*mol^-1)=0.3183*mol$ And thus mass of carbon...$-=0.3183*molxx12.011*g*mol^-1=3.824*g$ $"Moles...

1 Answers 1 views
)A compound that contains just C, H and N is burned in oxygen. When 5.000 g of the compound is burned in excess oxygen, 6.019 g of $CO_2$, 4.313 g of $H_2O$, and 9.438g of $NO_2$ are the only products. What is the empirical formula of the compound?

We must calculate the masses of $"C, H"$, and $"N"$ from the masses of carbon dioxide, water, and nitrogen dioxide. $"Mass of C" = 6.019 color(red)(cancel(color(black)("g CO"_2))) × "12.01 g...

1 Answers 1 views
X