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slava [35]
2 years ago
9

How many moles of carbon monoxide are there in 36. 55 g of carbon monoxide?

SAT
1 answer:
Natasha_Volkova [10]2 years ago
6 0

Answer:

mass of carbon monoxide

m = 36.55g

The chemical formula of carbon monoxide is,

C

O

.

Calculating the molar mass of carbon monoxide:

M

=

1

×

12

g

m

o

l

+

1

×

16

g

m

o

l

=

28

g

m

o

l

Calculating the number of moles:

n

=

m

M

=

36.55

g

28

g

m

o

l

=

1.305

m

o

l

Therefore, the are

1.305

moles in

36.55

grams of carbon monoxide

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Determine the empirical formula of a compound containing 47. 37 grams of carbon, 10. 59 grams of hydrogen, and 42. 04 grams of o
Effectus [21]

Considering the definition of empirical formula, the empirical formula is C₃H₈O₂ and the molecular formula is C₉H₂₄O₆.

<h3>Definition of empirical formula</h3>

The empirical formula is the simplest expression to represent a chemical compound, which indicates the elements that are present and the minimum proportion in whole numbers that exist between its atoms, that is, the subscripts of chemical formulas are reduced to the most integers. small as possible.

<h3>Empirical formula in this case</h3>

In this case, you have:

  • Carbon (C): 47.37 grams
  • Hydrogen (H): 10.59 grams
  • Oxygen (O):  42.04 grams

Then it is possible to calculate the number of moles of each atom in the molecule, taking into account the corresponding molar mass:

  • Carbon (C): \frac{47.37 grams}{12 \frac{g}{mole} }= 3.95 moles
  • Hydrogen (H): \frac{10.59 grams}{1 \frac{g}{mole} }= 10.59 moles
  • Oxygen (O):  \frac{42.04 grams}{16 \frac{g}{mole} }= 2.63 moles

The empirical formula must be expressed using whole number relationships, for this the numbers of moles are divided by the smallest result of those obtained. In this case:

  • Carbon (C): \frac{3.95 moles}{2.63 moles }= 1.5
  • Hydrogen (H): \frac{10.59 moles}{2.63 moles }= 4
  • Oxygen (O):  \frac{2.63 moles}{2.63 moles }= 1

Finally, since in the empirical formula the numbers of moles must be expressed in whole numbers, the ratio of atoms is multiplied by some number with which everything is obtained in simple whole numbers (in this case by 2):

  • Carbon (C): 1.5× 2= 3
  • Hydrogen (H): 4× 2= 8
  • Oxygen (O):  1× 2= 2

Therefore the C: H: O mole ratio is 3: 8: 2

Finally, the empirical formula is C₃H₈O₂.

<h3>Molecular formula</h3>

To obtain the molecular formula you must relate its molecular weight (PMc) with the molecular weight of the empirical formula (PMfe).

PMfe= PM (C₃H₈O₂)= 76 g/mole

The molar mass of the compound (PMc) was determined to be 228.276 g/mol. Dividing this mass PMc by the molar mass of the empirical formula PMfe gives:

228.276 g/mol ÷ 76 g/mole= 3

This means that in the molecular formula there are 3 unit formulas (empirical formulas), so it is necessary to multiply the number of all atoms by 3.

Then, the molecular formula is C₉H₂₄O₆.

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