What is the empirical formula for a substance that contains 25.94% nitrogen and 74.06% oxygen?

EMPIRICAL FORMULA
•The empirical formula of a compound gives the simplest whole number ratio of the atoms or ions present in the compound and can be found only by experiment.•The ratio of elements by mass must be converted to ratio of elements by numbers (amount of substance).Steps in calculating an empirical formula:1.Write down elements symbols.2.Calculate moles (n) for all elements.3.Divide all of the moles by the smallest number of moles.4.If necessary, multiply by a factor to convert all numbers to whole numbers.

Example:

A compound of sulfur contains 2.4 % hydrogen, 39.0 % sulfur and 58.6 % oxygen. Find the empirical formula of the compound.

Solution:

n (H) : n (S) : n (0)

2.4 : 39.0 : 58.6

1 32 162.4 : 1.2 : 3.7(Divide by smallest number – 1.2)

2.4 : 1.2 : 3.7

1.2 1.2 1.22 : 1 : 3 Empirical Formula is H2SO3

Empirical formula of a hydrated salt

•A hydrated compound is an ionic compound that produces water when heated.•An example of a hydrated salt is copper sulfate pentahydrate. CuSO4.5H20•When calculating the empirical formula of a hydrated compound you are finding the number of water molecules in the formula.

Example:

5.00 g of sodium carbonate were dried and 1.85 g remained. Find the empirical formula of the hydrated compound.

Solution:

n (Na2CO3) : n (H2O)

1.85 : 3.15

106 18 0.0175 : 0.175 (Divide by smallest number – 0.0175)

0.0175 : 0.175

0.0175 0.0175 1 : 10 Empirical Formula is Na2CO3 . 10H2O

MOLECULAR FORMULA

•The molecular formula of a compound represents the actual composition of a compound that is made up of molecules.•A molecular formula is either the same as the empirical formula, or is a whole number multiple of it.Example:The Empirical formula of hydrazine is NH2. The compound has a Molar Mass of 32g mol^-1. What is its molecular formula?M(NH2) = 16 g mol^-1- this is twice the molar mass of the molecule- 16x2 = 32 g mol^-1. So the molecular formula is double the empirical formula- N2H4.

PERCENTAGE COMPOSITION

•The percentage composition of a compound expresses in terms of percentage what element contributes to its overall mass.·To calculate percentage composition do the following:

% of X in compound containing X = mass of X in compound x 100
Molar mass of compound 1

Example:

Find the % composition by mass of each element in stannous fluoride, SnF2.% Sn= (mass (Sn) / M (SnF2) )x 100= (1 x 118.7 / 156.7) x 100= 75.7 %% F= (mass (F) / M (SnF2) )x 100= (2 x 19 / 156.7) x 100= 24.3 %

The formula for a hydrate quiz

Empirical and molecular formula quiz

Kahoot quiz- mole, empirical and molecular formula

Answer

Verified

Hint: Molecular mass is $2\times $Vapour density. Molecular formula is whole number multiple of empirical formula. Empirical formula is the simplest whole number ratio of various atoms present in a compound. mass % of an element is the ratio of mass of that element in compound to molar mass of compound.

Complete solution step by step:

-Empirical formula is the simplest whole number ratio of various atoms present in a compound.-Molecular formula shows exact number of different types of atoms present in a molecule of a compound.-Empirical formula can be calculated If percent mass of element is known. If molar mass is known, molecular formula can be calculated.- A compound contains N=25.94% and O=74.06%. so 100g of compound will contain 25.94g of N and 74.06 g of O.-As atomic mass of nitrogen is 14 g per mole and atomic mass of oxygen is 16 g per mole.Number of moles of N= $=\dfrac{25.94}{14}=1.85$Number of moles of O= $\dfrac{74.06}{16}=4.63$The mole ratio of nitrogen to oxygen is 2: 5 as numbers can be rounded off to make whole numbers.Hence, the empirical formula is ${{N}_{2}}{{O}_{5}}$. Empirical mass =28+80=108gMolecular mass= $2\times $vapour density=108.4Molecular formula= $n\times $ Empirical formula$n=\dfrac{molecular\text{ mass}}{empirical\text{ mass}}=\dfrac{108.4}{108}=1$Hence, Molecular formula=Empirical formulaA compound contains N=25.94% and O=74.06%. If its vapour density is 54.2, its molecular formula is (D) ${{N}_{2}}{{O}_{5}}$

Note: molecular formula can be determined also by this method. As we know molecular mass is twice of vapour density. Hence Molecular mass is approximately 108g, only option (D) has molecular mass as 108g. Empirical formula is the simplest whole number ratio of various atoms present in a compound. mass % of an element is the ratio of mass of that element in compound to molar mass of compound.