Answer:
0.0300 moles of H₂
Explanation:
The original equation is PV = nRT. We need to change this to show moles (n).
n =
It's important to convert your values to match the constant (r) in terms of units.
30.0 kPa = 0.296 atm
2500 mL = 2.50 L
27 °C = 300 K
Now, plug those values in to solve:
n = - for the sake of keeping the problem clean, I didn't include the units but you should just to make sure everything cancels out :)
Finally, you are left with n = 0.0300 moles of H₂
Answer: The empirical formula for the given compound is
Explanation:
The chemical equation for the combustion of compound having carbon, hydrogen and oxygen follows:
where, 'x', 'y' and 'z' are the subscripts of carbon, hydrogen and oxygen respectively.
We are given:
Mass of
Mass of
We know that:
Molar mass of carbon dioxide = 44 g/mol
Molar mass of water = 18 g/mol
For calculating the mass of carbon:
In 44 g of carbon dioxide, 12 g of carbon is contained.
So, in 28.6 g of carbon dioxide, of carbon will be contained.
For calculating the mass of hydrogen:
In 18 g of water, 2 g of hydrogen is contained.
So, in 14.6 g of water, of hydrogen will be contained.
For calculating the mass of oxygen:
Mass of oxygen in the compound = (14.60) - (7.8 + 1.6) = 5.2 g
To formulate the empirical formula, we need to follow some steps:
Step 1: Converting the given masses into moles.
Moles of Carbon =
Moles of Hydrogen =
Moles of Oxygen =
Step 2: Calculating the mole ratio of the given elements.
For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.32 moles.
For Carbon =
For Hydrogen =
For Oxygen =
Step 3: Taking the mole ratio as their subscripts.
The ratio of C : H : O = 2 : 5 : 1
Hence, the empirical formula for the given compound is