Answer:
Explanation:
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In this case, according to the given chemical reaction, it is possible for us to calculate the produced grams of nitrogen monoxide by starting with 25.0 g of nitrogen via their 1:2 mole ratio and the molar masses of 30.1 g/mol and 28.02 g/mol, respectively and by some stoichiometry:
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Answer : The empirical formula of the compound is
Explanation :
The chemical equation for the combustion of hydrocarbon 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 1.043 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 0.5670 g of water, of hydrogen will be contained.
Mass of oxygen in the compound = (0.600) - (0.284 + 0.063) = 0.253 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 0.0158 moles.
For Carbon =
For Hydrogen =
For Oxygen =
The ratio of C : H : O = 1.5 : 4 : 1
To make in a whole number we are multiplying the ratio by 2, we get:
The ratio of C : H : O = 3 : 8 : 2
Step 3: Taking the mole ratio as their subscripts.
The ratio of C : H : O = 3 : 8 : 2
Hence, the empirical formula for the given compound is