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3 years ago

16. Formaldhyde has a mole ratio of 1 mole C: 2 moles H: 1 mole O. Its empirical formula is the same as its

Chemistry

1 answer:

Dominik [7]3 years ago

6 0

Answer:

d. 97.60 g

Explanation:

Given parameters:

Number of moles of formaldehyde = 3.25moles

Ratio:

C H O

1 2 1

Unknown:

Mass of this sample = ?

Solution:

The empirical formula of a compound is its simplest formula. It is the simplest whole number ratio of the atoms in a given substance.

The molecular formula is the actual formula of the compound.

Since the molecular and empirical formula are the same here, the formula of the compound is;

CH₂O

To find the mass of the formaldehyde, use the expression below;

Mass = number of moles x molar mass

molar mass of CH₂O = 12 + 2(1) + 16 = 30g/mol

Mass = 3.25 x 30 = 97.5g

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The combustion of 0.570 g of benzoic acid (ΔHcomb = 3,228 kJ/mol; MW = 122.12 g/mol) in a bomb calorimeter increased the tempera

torisob [31]

Answer:

The temperature change from the combustion of the glucose is 6.097°C.

Explanation:

Benzoic acid;

Enthaply of combustion of benzoic acid = 3,228 kJ/mol

Mass of benzoic acid = 0.570 g

Moles of benzoic acid = $\frac{0.570 g}{122.12 g/mol}=0.004667 mol$

Energy released by 0.004667 moles of benzoic acid on combustion:

$Q=3,228 kJ/mol \times 0.004667 mol=15.0668 kJ=15,066.8 J$

Heat capacity of the calorimeter = C

Change in temperature of the calorimeter = ΔT = 2.053°C

$Q=C\times \Delta T$

$15,066.8 J=C\times 2.053^oC$

$C=7,338.92 J/^oC$

Glucose:

Enthaply of combustion of glucose= 2,780 kJ/mol.

Mass of glucose=2.900 g

Moles of glucose = $\frac{2.900 g}{180.16 g/mol}=0.016097 mol$

Energy released by the 0.016097 moles of calorimeter combustion:

$Q'=2,780 kJ/mol \times 0.016097 mol=44.7491 kJ=44,749.1 J$

Heat capacity of the calorimeter = C (calculated above)

Change in temperature of the calorimeter on combustion of glucose = ΔT'

$Q'=C\times \Delta T'$

$44,749.1 J=7,338.92 J/^oC\times \Delta T'$

$\Delta T'=6.097^oC$

The temperature change from the combustion of the glucose is 6.097°C.

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Answer:

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