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IceJOKER [234]
3 years ago
11

An 84-mg sample of a compound is found to contain 36 mg of carbon, 3 mg of hydrogen, 21 mg of nitrogen, and 24 mg of oxygen. If

the compound has a molecular weight of 112 g/mol, what is its molecular formula?
Chemistry
1 answer:
ANTONII [103]3 years ago
5 0

Answer:

The molecular formula of this compound is C4H4N2O2

Explanation:

Step 1: Data given

Mass of the compound = 84 mg

The compound contains:

36 mg of Carbon

3 mg of hydrogen

21 mg of nitrogen

24 mg of oxygen

Molar mass of carbon = 12.01 g/mol

Molar mass of hydrogen = 1.01 g/mol

Molar mass of nitrogen = 14 g/mol

Molar mass of oxygen = 16 g/mol

Step 2: Calculate number of moles

Moles = mass / molar mass

Moles of carbon = 0.036 g/ 12.01 g/mol = 0.003 moles

Moles of hydrogen = 0.003 g/ 1.01 g/mol = 0.003 moles

Moles of nitrogen = 0.021 g/ 14 g/mol = 0.0015 moles

Moles of oxygen = 0.024 g/ 16 g/mol = 0.0015 moles

Step 3: Calculate mol ratio

We divide by the smallest amount of moles

C: 0.003 / 0.0015 = 2

H: 0.003 / 0.0015 = 2

N = 0.0015/0.0015 = 1

O = 0.0015/0.0015 = 1

The empirical formula is C2H2NO

The molecular mass of this empirical formula is 56 g/mol

Step 4: Calculate the molecular formula

We have to multiply the empirical formula by n

n = 112 g/mol / 56g/mol = 2

We have to multiply the empirical formula by 2

Molecular formula = 2*(C2H2NO) = C4H4N2O2

The molecular formula of this compound is C4H4N2O2

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Calculate the amount of heat transferred when 24.0g of CH3OH(g) is decomposed by this reaction at constant pressure.For a given
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Answer:

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Explanation:

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24,0 g of CH₃OH(g) are:

24,0g CH₃OH×\frac{1mol}{32,04186g} = 0,749 moles of CH₃OH(g)

As two moles of CH₃OH(g) transferred +252 kJ, 0,749 moles of CH₃OH(g) transferred:

0,749 molesCH₃OH×\frac{+252 kJ}{2moles} = <em>94,4 kJ</em>

<em></em>

When 2 moles of CH₄ are produced, the enthalpy change is +252 kJ. 82,6 kJ are:

82,6kJ×\frac{2moles}{252 kJ}= 0,656 moles of CH₄

0,656 moles of CH₄ are:

0,656 moles of CH₄×\frac{16,05 g}{1mol} = <em>10,5 g of CH₄</em>

<em></em>

38,7 g of CH₄(g) are:

38,7g CH×\frac{1mol}{16,05g} = 2,41 moles of CH₄(g)

As two moles of CH₄ released 252 kJ, 2,41 moles of CH₄(g) released:

2,41 moles CH₄×\frac{+252kJ}{2moles} = <em>304 kJ</em>

<em></em>

I hope it helps!

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