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TEA [102]
3 years ago
12

A sample of 1.000 g of a compound containing carbon and hydrogen reacts with oxygen at elevated temperature to yield 0.692g H2O

and 3.381g CO2.
a. Calculate the masses of C and H in the sample.
b. Does the compound contain any other elements
Chemistry
1 answer:
Kazeer [188]3 years ago
6 0

Answer:

A. 0.922g and 0.0768g

B. NO

Explanation:

A.

To know the answer, we need to find the mass of the carbon and hydrogen present and see if they add up.

Firstly we get the mass of the carbon and hydrogen.

For the carbon, we first divide by the molar mass of carbon iv oxide which is 44g/mol.

The number of moles of carbon iv oxide is thus 3.381/44 = 0.076841

Since there is only one carbon atom, the answer is same as the number of moles of carbon. The mass is the only different thing here. The mass is the number of moles multiplied by the atomic mass. The atomic mass of carbon is 12. Thus, the total mass would be 0.076840909090909 * 12 = 0.922g

Now we calculate for hydrogen. We first find the number of moles from that of water. The molar mass of water is 18g/mol.

The number of moles is thus 0.692/18 = 0.0384

Now since we have 2 atoms of hydrogen in one mole of water, the number of moles of water present is thus 0.0384 * 2 = 0.0768

Now, since we know the number of moles, we can get the mass. The atomic mass of hydrogen is 1. This means that the mass if the hydrogen in the compound is 0.0768 * 1 = 0.0768g

B. We simply add the masses of hydrogen to that of carbon and subtract from 1. This is 1 - 0.922 - 0.0768 = 0.0012

The answer is no. We only have that residue because of rounding up of values

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tigry1 [53]

Answer:

m = 0.122 m

Explanation:

Molality(m): A measure of the number of moles of solute in a solution corresponding to 1 kg or 1000 g of solvent.

\mathrm{Molality\:Formula}:\quad \mathrm m =  \dfrac{ mole\:of\:solute  }{ mass\:of\:solvent\:in\:kg  }

\mathrm m =  \dfrac{ mole}{ kg  }

\mathrm{Substitute\:the\:numbers\:into\:the\:"Molality\:Formula"}

\mathrm m =   \dfrac{ 0.250  }{ 2.05  }

\mathrm{Divide\:0.250\:by\:2.05\:and\:get\:0.12195122}

\mathrm m =   {0.12195122}

\mathrm{Round\:to\:the\:nearest\:thousandths\:place}

\mathrm m =   0.122

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