A sample of ethanol (C2H6O) has a mass of
1 answer:
Answer:
-6.517 kJ is the heat of combustion for this sample that is 0.2301 grams of ethanol.
Explanation:
Mass of bomb calorimeter = 2.000 kg = 2.000 × 1000 g = 2000. g
1 kg = 1000 g
Specific heat of bomb calorimeter = c = 2.45 J/g°C
Change in temperature =
Heat absorbed by the bomb claorimeter = Q
Heat energy released on combustion of 0.2301 grams of ethanol = Q'= -Q
Q' = -Q = -6,517 J = -6.517 kJ
1 J = 0.001 kJ
-6.517 kJ is the heat of combustion for this sample that is 0.2301 grams of ethanol.
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Answer: The molecular formula for the given organic compound X is
Explanation:
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 44g of carbon dioxide, 12 g of carbon is contained.
So, in 4.13 g of carbon dioxide, = of carbon will be contained.
For calculating the mass of hydrogen:
In 18g of water, 2 g of hydrogen is contained.
So, in 1.13 g of water, of hydrogen will be contained.
Mass of oxygen in the compound = (3.00) - (1.13+ 0.125) = 1.75 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
For Carbon =
For Hydrogen =
For Oxygen =
Step 3: Taking the mole ratio as their subscripts.
The ratio of C : H : O = 1: 1.33: 1.16
Converting them into whole number ratios by multiplying by 6:
The ratio of C : H : O = 6: 8: 7
Hence, the empirical formula for the given compound is
Empirical mass =
The equation used to calculate the valency is :
Putting values in above equation, we get:
Multiplying this valency by the subscript of every element of empirical formula, we get:
Thus molecular formula for the given organic compound X is