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Bingel [31]
3 years ago
5

if 0.3 moles of carbon reacts completely with 0.3 moles of molecular oxygen, what is the yield in Percent of 6.6 g of CO2 are fo

rmed
Chemistry
1 answer:
Anettt [7]3 years ago
8 0
Answer is:<span>the yield is 50%.
</span>
Chemical reaction: C + O₂ → CO₂.
n(C) = 0.3 mol; amount of substance.
n(O₂) = 0.3 mol.
From chemical reaction: n(C) : n(CO₂) = 1 : 1.
n(CO₂) = 0.3 mol.
M(CO₂) = 44 g/mol; molar mass of caron(IV) oxide.
m(CO₂) = n(CO₂) · M(CO₂).
m(CO₂) =0.3 mol · 44 g/mol.
m(CO₂) = 13.2 g; mass of carbon(IV) oxide.
the yield = 6.6 g ÷ 13.2 g · 100%.
the yield = 50%.
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We have the next % composition:

C. 49.48%

H, 5.19%.

N. 28.85%

0. 16.48%

We assume 100 g of sample

1) As we have 100 g of sample of Caffeine, we calculate the mass of each element involved here.

C. 49.48 g

H, 5.19 g

N. 28.85 g

0. 16.48 g

2) We calculate the number of moles of each element (we need the mass per mole of each element)

For C) 12.01 g/mol

49.48 g x (1 mol/12.01 g) = 4.120 moles

For H) 1.007 g/mol

5.19 g x (1 mol/1.007 g) = 5.154 moles

For O) 15.99 g/mol

16.48 g x (1 mol/15.99 g) = 1.030 moles

For N) 14.00 g/mol

28.85 g x (1 mol/14.00 g) = 2.060 moles

3) We choose the smallest number from 2) and divide the rest of them by it.

For C) 4.120 moles/1.030 moles= 4

For H) 5.154 moles/1.030 moles= 5

For O) 1.030 moles/1.030 moles= 1

For N) 2.060 moles/1.030 moles= 2

4) The numbers in 3) represents the subindex from the empirical formula of caffeine:

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5) We calculate the molar mass of our empirical formula, 97.06 g/mol.

We already have the molar mass of the molecular formula, so we proceed like this:

n= the molar mass of the molecular formula/the molar mass of the empirical formula

n = 194.19 g/mol/97.06 g/mol = 2 approx.

We use "n" and we multiply our empirical formula by n = 2:

Therefore, our molecular formula:

C_8H_{10}O_2N_4

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