Burning a compound of calcium, carbon, and nitrogen in oxygen in a combustion train generates calcium oxide (CaO), carbon dioxide $(CO_2)$ , nitrogen dioxide $(NO_2)$ , and no other substances. A small sample gives 2.389 g CaO, 1.876 g $CO_2$ , and 3.921 g $NO_2$ Determine the empirical formula of the compound.

Answer: The empirical formula for the given compound is $CaCN_2$

Explanation:

The chemical equation for the combustion of compound having calcium, carbon and nitrogen follows:

$Ca_xC_yN_z+O_2\rightarrow CaO+CO_2+NO_2$

where, 'x', 'y' and 'z' are the subscripts of calcium, carbon and nitrogen respectively.

We are given:

Mass of CaO = 2.389 g

Mass of $CO_2=1.876g$

Mass of $NO_2=3.921g$

We know that:

Molar mass of calcium oxide = 56 g/mol

Molar mass of carbon dioxide = 44 g/mol

Molar mass of nitrogen dioxide = 46 g/mol

For calculating the mass of carbon:

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

So, in 1.876 g of carbon dioxide, $\frac{12}{44}\times 1.876=0.5116g$ of carbon will be contained.

For calculating the mass of nitrogen:

In 46 g of nitrogen dioxide, 14 g of nitrogen is contained.

So, in 3.921 g of nitrogen dioxide, $\frac{14}{46}\times 3.921=1.193g$ of nitrogen will be contained.

For calculating the mass of calcium:

In 56 g of calcium oxide, 40 g of calcium is contained.

So, in 2.389 g of calcium oxide, $\frac{40}{56}\times 2.389=1.706g$ of calcium will be contained.

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

Step 1: Converting the given masses into moles.

Moles of Calcium = $\frac{\text{Given mass of Calcium}}{\text{Molar mass of Calcium}}=\frac{1.706g}{40g/mole}=0.0426moles$

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.5116g}{12g/mole}=0.0426moles$

Moles of Nitrogen = $\frac{\text{Given mass of Nitrogen}}{\text{Molar mass of Nitrogen}}=\frac{1.193g}{14g/mole}=0.0852moles$

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 0.0426 moles.

For Calcium = $\frac{0.0426}{0.0426}=1$

For Carbon = $\frac{0.0426}{0.0426}=1$

For Nitrogen = $\frac{0.0852}{0.0426}=2$

Step 3: Taking the mole ratio as their subscripts.

The ratio of Ca : C : N = 1 : 1 : 2

Hence, the empirical formula for the given compound is $CaCN_2$

3 0

3 years ago

Imagine that you have a 5.00 L gas tank and a 3.50 L gas tank. You need to fill one tank with oxygen and the other with acetylen

Lorico [155]

Answer:

77.14 atm of pressure should be of an acetylene in the tank.

Explanation:

$2C_2H_2+5O_2\rightarrow 4CO_2+2H_2O$

According to reaction, 2 moles of acetylene reacts with 5 moles of oxygen.

Moles of oxygen= $n_1$

Moles of acetylene = $n_2$

$\frac{n_1}{n_2}=\frac{5 mol}{2 mol}=\frac{5}{2}$

Volume of large tank with oxygen gas, $V_1 = 5.00 L$

Pressure of the oxygen gas inside the tank = $P_1=135 atm$

$RT=\frac{P_1V_1}{n_1}$ ..[1]

Volume of small tank with acetylene gas , $V_2= 3.50 L$

Pressure of the acetylene gas inside the tank = $P_2=?$

$RT=\frac{P_2V_1}{n_2}$ ..[2]

Considering both the gases having same temperature T, [1]=[2]

$\frac{P_1V_1}{n_1}=\frac{P_2V_2}{n_2}$

$P_2=\frac{P_1V_1\times n_2}{V_2\times n_1}$

$=\frac{135 atm\times 5.00 L\times 2}{3.50 L\times 5}=77.14 atm$

77.14 atm of pressure should be of an acetylene in the tank.

8 0

3 years ago

Air is a _______. a. homogeneous mixture c. compound of oxygen and nitrogen b. heterogeneous mixture d. pure substance

SVEN [57.7K]

a because it was a combination of lots of subtances like CO2, oxygen, nitrogen,... and u cant see air so it is a homogenous mixture

7 0

3 years ago

2.68 x 10^24 atoms of Cu equal how many moles? Help!!!

Scilla [17]

Answer:- $2.68*10^2^4$ atoms of Cu are equals to 4.45 moles.

Solution:- It is a atoms to moles unit conversion for which we divide the given number of atoms by Avogadro number.

$1 mol = 6.02*10^2^3 atoms$

We have been given with $2.68*10^2^4$ atoms and asked to convert them to moles.

The set up for this unit conversion problem would be as:

$2.68*10^2^4 atoms(\frac{1mol}{6.02*10^2^3 atoms})$

= 4.45 moles

So, there would be 4.45 moles of Cu in $2.68*10^2^4$ atoms of Cu.

4 0

3 years ago

Read 2 more answers