mauricioalyssamarie
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a carefully weighed sample of compound (x)is burned which produced CO² and H²O that are measured accurately

Sagot :

mhiles4985

Answer:

The analysis showed the following percentage composition by mass: The molar mass is 60 g/mol. Determine its (a) empirical formula and (b) molecular mass

C = 40,0% H = 6,73% 0 = 53.3%

What is the empirical formula of the compound?

Solution:

A. In 100g of compound X

mass of C = 40.0 g

mass of H = 6.73 g

mass of O = 53.3g

B. You need to know the atomic mass of each element to solve this problem. These are:

atomic mass: C = 12 H = 1 O = 16

C. Get the ration in number of moles by dividing the mass of each element by its molar mass (this is equal to the atomic mass of the s = element).

C => 40.0 g = 3.33 moles

12 g / mol

H => 6.73 g = 6.73 moles

1 g / mol

O => 53.3 g = 3.33 moles

16 g / mol

D. The formula should express the smallest whole number ratio of atoms. These are expressed by the subscripts written for each symbol. To obtain this ratio divide the calculated number of moles by the least common denominator, 3 33 in this case.

C: 3.33 = 1 H: 5.73 = 2 O: 3.33 = 1

3.33 3.33 3.33

Thus, the empirical formula of the compound is CH₂O. There are several compounds, which this empirical formula. These are ethanoic acid or acetic acid (C2H4O2), glucose (C6H12O6), methanal or formaldehyde (HCHO), ribose (C5H10O5), lactic acid (C3H6O3) and many more.

To solve for the molecular formula:

A. Determine the empirical mass based on the empirical formula. This is the sum of the mass of all empirical mass

atoms contained in the empirical formula. Multiply the number of atoms present by its atomic mass to get the mass of each element present

Empirical mass:

(g) = (1 x 12g C) + (2x 1g H) + (1x 16g 0) = 30g

B. Find a factor F which when multiplied by the empirical mass will give the molar mass. To obtain this, simply divide the molar mass by the empirical mass.

F = 60/30=2

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