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yulyashka [42]
3 years ago
15

Combustion analysis of 0.300 g of an unknown compound containing carbon, hydrogen, and oxygen produced 0.5213 g of co2 and 0.283

5 g of h2o. what is the empirical formula of the compound?
Chemistry
1 answer:
Lyrx [107]3 years ago
6 0
First, we have to get how many grams of C & H & O in the compound:
- the mass of C on CO2 = mass of CO2*molar mass of C /molar mass of CO2
                                        = 0.5213 * 12 / 44 = 0.142 g
- the mass of H atom on H2O = mass of H2O*molar mass of H / molar mass of H2O
                                                 =0.2835 * 2 / 18 = 0.0315 g
- the mass of O = the total mass - the mass of C atom - the mass of H atom
                          =  0.3 - 0.142 - 0.0315 = 0.1265 g
Convert the mass to mole by divided by molar mass
C(0.142/12) H(0.0315/2) O(0.1265/16) 
C(0.0118) H(0.01575) O(0.0079) by dividing by the smallest value 0.0079
C1.504 H3.99 O1 by rounding to the nearst fraction
C3/2 H4/1 )1/1 multiply by 2 
∴ the emprical formula C3H8O2
           



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Molar mass is the ratio of the mass to that amount of the substance. The mass of the barium nitrate in the formula unit is 23.0 grams.

<h3>What is mass?</h3>

The mass of a substance is the product of the molar mass of the compound and the number of moles of the compound.

Given,

Molar mass of barium nitrate = 261.35 g/mol

If, 6.022 \times 10^{23} have a mass of 261.35 g/mol then, 5.30 \times 10^{22} formula units will have a mass of,

\begin{aligned}& = \dfrac{261.35 \times 5.30 \times 10^{22}}{6.022\times 10^{23}}\\\\&= 23.0\;\rm gm\end{aligned}

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Learn more about mass here:

brainly.com/question/24958554

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