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ZanzabumX [31]
3 years ago
9

Convert 0.859 mg to cg. Help me plz :(

Chemistry
1 answer:
andrey2020 [161]3 years ago
7 0

Answer:

0.0859

Explanation:

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Compare endothermic and exothermic reactions and give examples of each. In an endothermic reaction would we expect the temperatu
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3 years ago
Combustion analysis of 0.600 g of an unknown compound containing carbon, hydrogen, and oxygen produced 1.043 g of CO2 and 0.5670
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Answer : The empirical formula of the compound is C_3H_8O_2

Explanation :

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

C_xH_yO_z+O_2\rightarrow CO_2+H_2O

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of CO_2=1.043g

Mass of H_2O=0.5670g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 1.043 g of carbon dioxide, \frac{12}{44}\times 1.043=0.284g of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 0.5670 g of water, \frac{2}{18}\times 0.5670=0.063g of hydrogen will be contained.

Mass of oxygen in the compound = (0.600) - (0.284 + 0.063) = 0.253 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.284g}{12g/mole}=0.0237moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.063g}{1g/mole}=0.063moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.253g}{16g/mole}=0.0158moles

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.0158 moles.

For Carbon = \frac{0.0237}{0.0158}=1.5

For Hydrogen  = \frac{0.063}{0.0158}=3.98\approx 4

For Oxygen  = \frac{0.0158}{0.0158}=1

The ratio of C : H : O = 1.5 : 4 : 1

To make in a whole number we are multiplying the ratio by 2, we get:

The ratio of C : H : O = 3 : 8 : 2

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 3 : 8 : 2

Hence, the empirical formula for the given compound is C_3H_8O_2

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