The graph shows the amount of a gaseous product formed over time during two trials of a reaction. A different concentration of a
reactant was used during each trial while other factors were kept constant. (see graph)
Which of the following statements explains which trial has a higher concentration of the reactant?
Trial 1, because the final amount of product formed is higher.
Trial 1, because this reaction lasted for a longer duration than Trial 2.
Trial 2, because this reaction was initially fast and slowed down later.
Trial 2, because the amount of product formed per unit time is higher.
Which of the following statements explains which trial has a higher concentration of the reactant?
Trial 1, because the final amount of product formed is higher.
Trial 1, because this reaction lasted for a longer duration than Trial 2.
Trial 2, because this reaction was initially fast and slowed down later.
Trial 2, because the amount of product formed per unit time is higher.
2 answers:
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<u>Answer:</u> The empirical and molecular formula for the given organic compound is and
<u>Explanation:</u>
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
- <u>For calculating the mass of carbon:</u>
In 44g of carbon dioxide, 12 g of carbon is contained.
So, in 9.06 g of carbon dioxide, of carbon will be contained.
- <u>For calculating the mass of hydrogen:</u>
In 18g of water, 2 g of hydrogen is contained.
So, in 5.58 g of water, of hydrogen will be contained.
- Mass of oxygen in the compound = (6.38) - (2.47 + 0.62) = 3.29 g
To formulate the empirical formula, we need to follow some steps:
- <u>Step 1:</u> Converting the given masses into moles.
Moles of Carbon =
Moles of Hydrogen =
Moles of Oxygen =
- <u>Step 2:</u> 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 0.206 moles.
For Carbon =
For Hydrogen =
For Oxygen =
- <u>Step 3:</u> Taking the mole ratio as their subscripts.
The ratio of C : H : O = 1 : 3 : 1
Hence, the empirical formula for the given compound is
For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.
The equation used to calculate the valency is:
We are given:
Mass of molecular formula = 124 amu = 124 g/mol
Mass of empirical formula = 31 g/mol
Putting values in above equation, we get:
Multiplying this valency by the subscript of every element of empirical formula, we get:
Thus, the empirical and molecular formula for the given organic compound is and