2.4g of a compound of carbon, hydrogen and oxygen gave on combustion, 3.52g of CO2 and 1.44g of H2O. The relative molecular mass of the compound was found to be 60.

a)What are the masses of carbon, hydrogen and oxygen in 2.4g of the compound?

b)What are the emperical and molecular formulae of the compound?

2.5g of a compound of carbon hydrogen and water gave on combustion 3.52g of carbon dioxyde and 1.44g of water. Thé relative molecular mass of the compound 180. Determine imperical and molecular formulae of the compound

a.

3.52g CO2 x (atomic mass C/molar mass CO2) = g C.
1.44 g H2O x (2*atomic mass H/molar mass H2O) = g H.
g O = 2.4 - gC - g H

b. Can you finish? If you need more assistance please post your work and explain what you don't understand in detail.

I didn't understand anything

1. 2.4g of a compound X consisting of carbon, oxygen and hydrogen gave on

combustion 3,52g of CO2 and 1.44g of H2O
a. What are the masses of carbon, oxygen and hydrogen in the 2.4g of
compound X?
b. What is the empirical formula of compound X
c. If molar mass of X is 60g, what is its molecular formula
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thank you very much Dr.bob222

a) To find the masses of carbon, hydrogen, and oxygen in 2.4g of the compound, we need to calculate the moles of CO2 and H2O produced during combustion.

Moles of CO2 = mass of CO2 / molar mass of CO2
Moles of CO2 = 3.52g / 44g/mol (molar mass of CO2)
Moles of CO2 ≈ 0.08 mol

Moles of H2O = mass of H2O / molar mass of H2O
Moles of H2O = 1.44g / 18g/mol (molar mass of H2O)
Moles of H2O ≈ 0.08 mol

Since the empirical formula gives the simplest whole number ratio of elements, we divide the number of moles of each element by the smallest number of moles.

Moles of C = 0.08 mol / 0.08 mol = 1 mole
Moles of H = 0.08 mol / 0.08 mol = 1 mole
Moles of O = 0.08 mol / 0.08 mol = 1 mole

To find the masses of each element, we multiply their respective moles by their molar masses.

Mass of C = 1 mol × 12 g/mol (molar mass of C) = 12g
Mass of H = 1 mol × 1 g/mol (molar mass of H) = 1g
Mass of O = 1 mol × 16 g/mol (molar mass of O) = 16g

Therefore, the masses of carbon, hydrogen, and oxygen in 2.4g of the compound are approximately:
Carbon: 12g
Hydrogen: 1g
Oxygen: 16g

b) The empirical formula of the compound can be determined by dividing the moles of each element by the smallest number of moles, which is 0.08 mol for each element. Since all moles are the same, the empirical formula is CH2O.

To find the molecular formula, we need to compare the empirical formula mass (EFM) to the given relative molecular mass (RMM) of 60.

EFM of CH2O = (12g/mol × 1) + (1g/mol × 2) + (16g/mol × 1)
EFM of CH2O = 12g + 2g + 16g = 30g/mol

The molecular formula mass should be a multiple of the empirical formula mass. We divide the RMM by the EFM to find the ratio.

Molecular formula ratio = RMM / EFM
Molecular formula ratio = 60 / 30
Molecular formula ratio = 2

Therefore, the molecular formula is two times the empirical formula:
C2H4O2

To solve this problem, we need to use the concept of stoichiometry and the law of conservation of mass.

a) To find the masses of carbon, hydrogen, and oxygen in 2.4g of the compound, we can follow these steps:
1. Calculate the mass of CO2 produced:
- From the given information, we know that 3.52g of CO2 is produced during combustion.
- Since the molar mass of CO2 is 44 g/mol, we can calculate the number of moles using the equation: moles = mass / molar mass.
moles of CO2 = 3.52g / 44 g/mol = 0.08 mol.

2. Calculate the mass of H2O produced:
- From the given information, we know that 1.44g of H2O is produced during combustion.
- Since the molar mass of H2O is 18 g/mol, we can calculate the number of moles using the equation: moles = mass / molar mass.
moles of H2O = 1.44g / 18 g/mol = 0.08 mol.

3. Next, we need to determine the number of moles of carbon, hydrogen, and oxygen in the compound.
- Carbon is present in CO2, and from the balanced equation for the combustion, we know that each mole of CO2 contains one mole of carbon.
moles of carbon = 0.08 mol.

- Hydrogen is present in H2O, and from the balanced equation for the combustion, we know that each mole of H2O contains two moles of hydrogen.
moles of hydrogen = 2 * 0.08 mol = 0.16 mol.

- To find the moles of oxygen, we can subtract the sum of the moles of carbon and hydrogen from the total moles of the compound.
moles of oxygen = Total moles - moles of carbon - moles of hydrogen
= 0.24 mol - 0.08 mol - 0.16 mol
= 0 mol (since the total moles of the compound are not given).

4. Calculate the masses of carbon, hydrogen, and oxygen in 2.4g of the compound using their molar masses:
- The relative molecular mass of the compound is given as 60, which means that the sum of the molar masses of carbon, hydrogen, and oxygen is equal to 60 g/mol.
- Since we know the moles of carbon and hydrogen (0.08 mol each), we can calculate the moles of oxygen:
moles of oxygen = 60 g/mol - (moles of carbon + moles of hydrogen)
= 60 g/mol - (0.08 mol + 0.16 mol)
= 60 g/mol - 0.24 mol
= 59.76 g/mol.

- Finally, calculate the masses of carbon, hydrogen, and oxygen using their moles and molar masses:
mass of carbon = moles of carbon * molar mass of carbon
= 0.08 mol * atomic mass of carbon

mass of hydrogen = moles of hydrogen * molar mass of hydrogen
= 0.16 mol * atomic mass of hydrogen

mass of oxygen = moles of oxygen * molar mass of oxygen
= 59.76 g/mol * atomic mass of oxygen

b) To determine the empirical formula and molecular formula of the compound, we need the molar ratio between carbon, hydrogen, and oxygen.
- Using the moles calculated in part a), we can find the simplest ratio by dividing each mole value by the smallest mole value.
- Multiply all the values by a factor, if needed, to obtain whole numbers.
- The empirical formula is the simplest ratio of atoms in a compound.

To find the molecular formula of the compound, we need to know the molar mass of the compound.
If the molar mass of the compound is a multiple of the empirical formula mass, it means that the empirical formula is the same as the molecular formula.
If the molar mass is not a multiple, it means that the molecular formula is a multiple of the empirical formula.

Given that the relative molecular mass of the compound is 60 g/mol, we can compare it to the empirical formula mass to determine the molecular formula.
If the empirical formula mass is equal to 60 g/mol, the molecular formula is the same as the empirical formula.
If the empirical formula mass is smaller than 60 g/mol, we need to find the multiple.

To calculate the empirical formula mass, we sum the atomic masses of the elements in the empirical formula and calculate the molar mass:
Empirical formula mass = (mass of carbon) + (mass of hydrogen) + (mass of oxygen)

Once we have the empirical formula mass, we can compare it to the molecular mass (60 g/mol) to determine the molecular formula by calculating the multiple.

I hope this explanation helps you better understand the steps involved in solving this problem!