problem 3.
A compound Y containing carbon,hydrogen and oxygen was burned in a stream of oxygen,the carbondioxide and water were cooled in a pre way absorption tube and carefully reweighed.Readings obtained were
initial mass of Y is 20.63mg
mass of carbondioxide produced 57.94mg
mass of water produced 11.85mg.
find the percentage of each element and hence the emperical formular of compound Y.
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To find the percentage of each element and the empirical formula of compound Y, we need to calculate the moles of carbon, hydrogen, and oxygen using the given information.
Step 1: Calculate the moles of carbon dioxide (CO2) and water (H2O) produced.
The molar mass of carbon dioxide (CO2) = 12.01 g/mol (carbon) + 2 * 16.00 g/mol (oxygens) = 44.01 g/mol
The moles of carbon dioxide produced = mass of carbon dioxide produced / molar mass of carbon dioxide
= 57.94 mg / 44.01 g/mol
= 1.3162 mmol
The molar mass of water (H2O) = 2 * 1.01 g/mol (hydrogens) + 16.00 g/mol (oxygen) = 18.02 g/mol
The moles of water produced = mass of water produced / molar mass of water
= 11.85 mg / 18.02 g/mol
= 0.6564 mmol
Step 2: Calculate the moles of carbon, hydrogen, and oxygen in compound Y.
In the combustion reaction, each mole of compound Y will produce one mole of carbon dioxide (CO2) and x moles of water (H2O), where x represents the number of moles of water per mole of compound Y.
From the stoichiometry of the combustion reaction, we can write the following equations:
1 mole of Y → 1 mole of CO2
1 mole of Y → x moles of H2O
Using the mole ratios, we can relate the moles of carbon dioxide and water to the moles of carbon, hydrogen, and oxygen in compound Y:
1.3162 mmol Y = 1.3162 mmol CO2 = 1.3162 mmol C
0.6564 mmol Y = x * 0.6564 mmol H2O = 2 * x * 0.6564 mmol H
Step 3: Calculate the mass of carbon, hydrogen, and oxygen in compound Y.
The mass of carbon in Y = moles of carbon in Y * atomic mass of carbon
= 1.3162 mmol * 12.01 g/mol
= 15.795 g
The mass of hydrogen in Y = moles of hydrogen in Y * atomic mass of hydrogen
= 2 * x * 0.6564 mmol * 1.01 g/mol
= 1.3305 g * x
The mass of oxygen in Y = mass of Y - mass of carbon - mass of hydrogen
= 20.63 mg - 15.795 g - 1.3305 g * x
= (20.63 - 15.795 - 1.3305x) g
Step 4: Calculate the percentage of each element in compound Y.
Percentage of carbon = (mass of carbon / mass of Y) * 100%
= (15.795 g / 20.63 mg) * 100%
= (15.795 / 20.63) * 100%
= 76.6%
Percentage of hydrogen = (mass of hydrogen / mass of Y) * 100%
= (1.3305 g * x / 20.63 mg) * 100%
= (1.3305 * x / 20.63) * 100%
= 6.45x%
Percentage of oxygen = (mass of oxygen / mass of Y) * 100%
= [(20.63 - 15.795 - 1.3305x) g / 20.63 mg] * 100%
We need to solve for x in the equation:
76.6% + 6.45x% + [(20.63 - 15.795 - 1.3305x) g / 20.63 mg] * 100% = 100%
Solving this equation will give us the value of x, which represents the number of moles of water per mole of compound Y.
Once we have the value of x, we can determine the empirical formula of compound Y. The empirical formula represents the simplest whole number ratio of atoms in a compound.