# Molecular Formuals from Empirical Formula

• Chemistry
• HigueraC
In summary, the unknown compound contains 21.8 g of phosphorus and 28.2 g of oxygen, with a molar mass of 284 g/mol. The empirical formula is PO3, and the molecular formula could be P2O5 or P4O10.
HigueraC

## Homework Statement

An unknown compound contains 21.8 g of phosphorus and 28.2 g of oxygen. It's molar mass is 284 g/mol. what are its empirical and molecular formulas?

Elements and the Weight:
P = 31 g/mol
O = 16 g/mol

Given Measurements:
21.8g P, 28.2g O, MM = 284g/mol

2. The attempt at a solution

This is my math:
Elements and the Weight:
P = 31 g/mol
O = 16 g/mol

Given Measurements:
21.8g P, 28.2g O, MM = 284g/mol

Converting the given measurements to moles:
21.8gP / 31 = .70 mol P
28.2gO / 16 = 1.8 mol O

Getting the ratio:
.70 / .70 = 1
1.8 / .70 = 2.57 (rounds up to 3)

Ratio:
1:3

Empirical Formula : PO3

Formula Mass of PO3:
79g/mol

Finding the common factor:
284(the given weight)/ 79(the weight of PO3) = 3.594 (rounds up to 4)

Writing the Molecular Formula:
4(PO3) = P4O12

Getting the Molar Mass of the Molecular Formula:
P4 = 124
O12 = 192
P4O12 = 316
316 =/= 284(is what it should be)

Nevermind, I did some research and found out that if the ratio is ~2.5 that you're supposed to multiply it by 2 to get even numbers.

HigueraC said:

## Homework Statement

An unknown compound contains 21.8 g of phosphorus and 28.2 g of oxygen. It's molar mass is 284 g/mol. what are its empirical and molecular formulas?

Elements and the Weight:
P = 31 g/mol
O = 16 g/mol

Given Measurements:
21.8g P, 28.2g O, MM = 284g/mol

2. The attempt at a solution

This is my math:
Elements and the Weight:
P = 31 g/mol
O = 16 g/mol

Given Measurements:
21.8g P, 28.2g O, MM = 284g/mol

Converting the given measurements to moles:
21.8gP / 31 = .70 mol P
28.2gO / 16 = 1.8 mol O

Getting the ratio:
.70 / .70 = 1
1.8 / .70 = 2.57 (rounds up to 3)

Ratio:
1:3

Empirical Formula : PO3

Formula Mass of PO3:
79g/mol

Finding the common factor:
284(the given weight)/ 79(the weight of PO3) = 3.594 (rounds up to 4)

Writing the Molecular Formula:
4(PO3) = P4O12

Getting the Molar Mass of the Molecular Formula:
P4 = 124
O12 = 192
P4O12 = 316
316 =/= 284(is what it should be)
====================
You got a ratio of 2.57:1. instead of rounding to 3, try 5:2, which is a lot closer.
Then your formula is P2O5 or P4O10. This makes sense since P and O have oxidation numbers of +5 and -2, so the electrons balance out to zero.

## 1. What is the difference between a molecular formula and an empirical formula?

A molecular formula shows the exact number of atoms of each element present in a molecule, while an empirical formula shows the simplest whole number ratio of atoms present in a compound.

## 2. How do you determine the molecular formula from the empirical formula?

The molecular formula can be determined by dividing the molecular mass of the compound by the empirical mass and then multiplying the subscripts in the empirical formula by this factor.

## 3. Can a compound have the same empirical and molecular formula?

Yes, a compound can have the same empirical and molecular formula if it is a simple molecule with the same ratio of atoms. However, for more complex molecules, the molecular formula will be a multiple of the empirical formula.

## 4. What information can be obtained from the molecular formula of a compound?

The molecular formula provides information about the types and number of atoms present in a molecule, which is essential for understanding the chemical properties and behavior of the compound.

## 5. Can the empirical formula be determined from the molecular formula?

Yes, the empirical formula can be determined from the molecular formula by finding the simplest whole number ratio of atoms present in the molecule.

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