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Eduardwww [97]
3 years ago
9

Measurements show that unknown compound has the following composition: element mass 62.1 % carbon, 10.5 % hydrogen and 27.6 % ox

ygen. Write the empirical chemical formula of this compound?
Chemistry
1 answer:
barxatty [35]3 years ago
8 0

Answer:

C3H6O

Explanation:

The percentage composition of the elements in the compound are given as follows:

62.1 % carbon = 62.1g of C

10.5 % hydrogen = 10.5g of H

27.6 % oxygen = 27.6g of O

Next, we convert each mass to mole by dividing by their molar/atomic mass

C = 62.1/12 = 5.175mol

H = 10.5/1 = 10.5mol

O = 27.6/16 = 1.725mol

Next, we divide each mole value by the smallest mole value (1.725)

C = 5.175mol ÷ 1.725 = 3

H = 10.5mol ÷ 1.725 = 6.086

O = 1.725mol ÷ 1.725 = 1

The empirical ratio approximately of C:H:O is 3:6:1, hence, the empirical formula is C3H6O

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\text{Number of moles}=\frac{71.0g}{63.01g/mol}=1.13moles

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<u>Answer:</u> The longest wavelength of light is 656.5 nm

<u>Explanation:</u>

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Putting the values in above equation, we get:

\frac{1}{\lambda }=1.096776\times 10^7m^{-1}\left(\frac{1}{2^2}-\frac{1}{3^2} \right )\\\\\lambda =\frac{1}{1.5233\times 10^6m^{-1}}=6.565\times 10^{-7}m

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