Chemistry, 03.08.2019 13:30 GreenHerbz206

Anitrogen oxide, containing 53.85% n, acts as a vasodilator, lowering blood pressure in the human body. what is its empirical formula? '

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The empirical formula of the nitrogen oxide that acts as a vasodilator and lowers human blood pressure is N_4O_3 . Empirical formula is the smallest whole number ratios of elements in a compound.

FURTHER EXPLANATION

To get the empirical formula from mass percent data, the following steps are followed:

1.Get the mass percent data for all the elements that make up the compound.

2.Assume that there is 100 grams of sample compound. Determine the equivalent masses of each element using the mass percent given.

3.Convert the mass of each element to number of moles.

4.Divide each calculated mole by the smallest mole value.

5.The quotient will be the subscript of the element in the compound. If a decimal is obtained, round off the answers to the nearest whole number. Some exceptions to this are the following decimals: x.25, x.33, x.50, and x.75. When these decimals are obtained, all the subscripts are multiplied by a number that will result in a whole number.

Applying the steps to the problem,

STEP 1: Get the mass percent.

Mass % of N = 53.85

Mass % of O = (100 - 53.85) = 46.15

STEP 2: Assume that 100 g sample is used and convert the mass % to mass in grams.

mass of N = 53.85% x 100 g = 53.85 g

mass of O= 46.15% x 100 g = 46.15 g

STEP 3: Convert mass to moles by dividing the given mass by the formula mass.

$moles \ of \ N \ = 53.85 \ g (\frac{1 \ mol}{14 \ g}) = 3.85 \\\\moles \ of \ O \ = 46.26 \ g \ (\frac{1 \ mol}{16 \ g}) \ = 2.89$

The moles may be written as the temporary subscripts of the elements in the compound as follows:

$N_{3.85}O_{2.89}$

STEP 4: Divide the moles by the smallest mole value.

$N_{3.85}O_{2.89}\\N_{ \frac{3.85}{2.89}} \ O _{ \frac{2.89}{2.89}} \\ N_{1.33}O_{1}\\$

STEP 5: Multiply the subscripts by a number that would give the smallest whole number ratio.

Since the decimal is 1.33 it cannot be rounded off to 1. It should be multiplied by 3 to get the smallest whole number.

NOTE: All subscripts must be multiplied by the same factor, 3.

$N_{1.33}O_{1}\\3(N_{1.33}O_{1})\\\boxed {N_{4}O_{3}}$

To check if the empirical formula is correct, calculate the mass % of each element based on the formula and compare with the given in the problem.

$mass \ \%\ = (\frac{mass \ of \ element}{mass \ of \ compound}) x 100\\\\mass \ \% \ of \ N = \ \frac{(4)(14)}{(4)(14) \ + \ (3)(16)}$

$mass \ \% \ of \ N \ = 53.85%\\mass \ \% \ of \ O \ = (\frac{(3)(16)}{(4)(14)+(3)(16)}) \ 100\\mass \ \% \ of \ O \ = 46.15%$

Since the values are the same from the given, the answer is correct.

LEARN MORE

Writing Chemical Formula Naming Compounds Mole Conversion

Keywords: empirical formula, compounds

Answer from: christopherschool04

The empirical formula of nitrogen oxide is $\boxed{{{\text{N}}_4}{{\text{O}}_3}}$ .

Further explanation:

Empirical formula:

It is atom’s simplest positive integer ratio in the compound. It may or may not be same as that of molecular formula. For example, empirical formula of sulfur dioxide is ${\text{SO}}$ .

Step 1: Mass percent of oxygen is to be calculated. This is done by using equation (1).

Since nitrogen oxide consists of only nitrogen and oxygen. So the percentage of oxygen is calculated as follows:

${\text{\% of O}}=100\;\% -{\text{\% of N}}$ …… (1)

The percentage of N is 53.85 %.

Substitute the value in equation (1).

$\begin{gathered}{\text{\% of O}}=100\;\% - 53.85\;{\text{\%}}\\={\text{46}}{\text{.15}}\;{\text{\% }}\\\end{gathered}$

Step 2: The mass of nitrogen and oxygen is to be calculated.

Consider the mass of nitrogen oxide to be 100 g.

The percent of N in nitrogen oxide is 53.85 %.

The mass of nitrogen oxide is 100 g.

The formula to calculate the mass of nitrogen is as follows:

${\text{Mass of N}}=\left( {\frac{{\%{\text{ of N}}}}{{100{\text{\%}}}}}\right)\left({{\text{Mass of nitrogen oxide}}}\right)$ …… (2)

Substitute these values in equation (1).

$\begin{aligned}{\text{Mass of N}}&=\left({\frac{{53.85\;\%}}{{100{\text{\%}}}}}\right)\left({100{\text{ g}}} \right)\\&={\text{53}}{\text{.85 g}}\\\end{aligned}$

The formula to calculate the mass of oxygen is as follows:

${\text{Mass of O}}=\left({\frac{{\% {\text{ of O}}}}{{100{\text{\%}}}}}\right)\left({{\text{Mass of nitrogen oxide}}}\right)$ …… (3)

The % of O in nitrogen oxide is 46.15 %.

The mass of nitrogen oxide is 100 g.

Substitute these values in equation (3).

$\begin{aligned}{\text{Mass of O}}&=\left({\frac{{46.15\;\%}}{{100{\text{\%}}}}}\right)\left({100{\text{g}}}\right)\\&={\text{46}}{\text{.15 g}}\\\end{aligned}$

Step 3: The moles of nitrogen and oxygen are to be calculated.

The formula to calculate the moles of nitrogen is as follows:

${\text{Moles of N}} = \frac{{{\text{Given mass of N}}}}{{{\text{Molar mass of N}}}}$ …… (5)

The given mass of O is 46.15 g.

The molar mass of O is 16 g/mol.

Substitute these values in equation (5).

$\begin{aligned}{\text{Moles of O}}&=\left({{\text{46}}{\text{.15 g}}}\right)\left({\frac{{{\text{1 mol}}}}{{{\text{16 g}}}}}\right)\\&=2.8843\\&\approx{\text{2}}{\text{.89 mol}}\\\end{aligned}$

Step 4: The moles of nitrogen and oxygen of nitrogen oxide are to be written with their corresponding subscripts in nitrogen oxide to get the preliminary formula.

So the preliminary formula becomes,

${\text{Preliminary formula of nitrogen oxide}} = {{\text{N}}_{3.85}}{{\text{O}}_{2.89}}$

Step 5: Each of the subscripts is divided by the smallest subscript to get the empirical formula.

In this case, the smallest one is 2.89. So the empirical formula of nitrogen oxide is written as follows:

$\begin{aligned}{\text{Empirical formula of nitrogen oxide}}={{\text{N}}_{\frac{{3.85}}{{2.89}}}}{{\text{O}}_{\frac{{2.89}}{{2.89}}}}\\={{\text{N}}_{{\text{1}}{\text{.33}}}}{{\text{O}}_{\text{1}}}\\\end{aligned}$

Step 6: Multiply each subscript of the empirical formula by 3 to get the whole number in subscript.

Therefore we get the final empirical formula as follows:

$\begin{aligned}{\text{Empirical formula of nitrogen oxide}}&={{\text{N}}_{3\left({{\text{1}}{\text{.33}}}\right)}}{{\text{O}}_{3\left({\text{1}}\right)}}\\&={{\text{N}}_{3.99}}{{\text{O}}_3}\\&\approx{{\text{N}}_4}{{\text{O}}_3}\\\end{aligned}$

Hence, the empirical formula is ${{\mathbf{N}}_{\mathbf{4}}}{{\mathbf{O}}_{\mathbf{3}}}$ .

Learn more:

1. Calculate the moles of ions in the solution:

2. Calculate the moles of chlorine in 8 moles of carbon tetrachloride:

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Stoichiometry of formulas and equations

Keywords: empirical formula, N4O3, N, O, nitrogen oxide, subscript, moles of nitrogen, moles of oxygen, mass of nitrogen, mass of oxygen.

Answer from: Quest

your answer is a. the enthalpy of formation for a pure element in its standard state is always positive.

Answer from: Quest

the answer is 0.8m.

by: nicholas

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