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kolezko [41]
3 years ago
10

Determine the empirical formula of a compound containing 47.37 grams of carbon, 10.59 grams of hydrogen, and 42.04 grams of oxyg

en.
In an experiment, the molar mass of the compound was determined to be 228.276 g/mol. What is the molecular formula of the compound?

For both questions, show your work or explain how you determined the formulas by giving specific values used in calculations. (10 points)
Chemistry
1 answer:
inysia [295]3 years ago
4 0

Answer:

Mass of C = 47.37g

Mass of H = 10.59g

Mass of O = 42.04g

The total mass of these elements is 100g, taking a proportion of their molar masses.

C = 47.37/12= 3.95

H = 10.59/1 = 10.59

O = 42.04/16= 2.63.

Dividing through with the smallest proportion which is 2.63

C=3.95/2.63 = 1.5

H =10.59/2.63 =4

O = 2.63/2.63= 1

Multiplying through by 2 to get a whole number.

C = 1.5x2 = 3

H= 4x2 = 8

O = 1x2= 2

The empirical formula is C3H6O2

(Empirical formula)n= molecular mass

(C3H8O2)n =228.276

(12x3 +8+16x2)n= 228.276

76n = 228.276

n = 228.276/76

n = 3

Molecular formula = Empirical formula

=(C3H8O2)3 = C9H24O6

The molecular formula is C9H24O6

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Explanation:

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5 0
3 years ago
2C2H2 + 5O2 → 4CO2 + 2 H2O How many grams of CO2 are required to react with exactly 3.00 mol of O2? 106 g 132 g 165 g 76.8 g
cricket20 [7]

Hey there!:

Given the reaction:

2 C2H2 + 5 O2 → 4 CO2 + 2 H2O

5 moles O2 ------------- 4 moles CO2

3.00 moles O2 ----------  ( moles of CO2 ?? )

moles of CO2 = 3.00 * 4 / 5

moles of CO2 =  12 / 5

moles of CO2 = 2.4 moles

So, molar  mass CO2 = 44.01 g/mol

Therefore:

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Answer A


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8 0
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timofeeve [1]

Answer:

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Answer:

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A '+' sign is written between the reactants, when more than one reactants are present.

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The products must be written on the right side of the direction arrow.

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For the given chemical equation:

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