What is the percent of hydrogen by mass in CH4O? (H:1.00 amu, C: 12.01 amu, O: 16.00 amu)

Chemistry

1 answer:

cricket20 [7]2 years ago

3 0

The mass percent of hydrogen in CH₄O is 12.5%.

<h3>What is the mass percent?</h3>

Mass percent is the mass of the element divided by the mass of the compound or solute.

Step 1: Calculate the mass of the compound.

mCH₄O = 1 mC + 4 mH + 1 mO = 1 (12.01 amu) + 4 (1.00 amu) + 1 (16.00 amu) = 32.01 amu

Step 2: Calculate the mass of hydrogen in the compound.

mH in mCH₄O = 4 mH = 4 (1.00 amu) = 4.00 amu

Step 3: Calculate the mass percent of hydrogen in the compound.

%H = (mH in mCH₄O / mCH₄O) × 100%

%H = 4.00 amu / 32.01 amu × 100% = 12.5%

The mass percent of hydrogen in CH₄O is 12.5%.

Learn more about mass percent here:brainly.com/question/4336659

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1) A mixture of anhydrous sodium carbonate and sodium hydrogencarbonate of mass 10.000 g was heated until it reached a constant

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The masses of the components are obtained as;

Sodium hydrogen carbonate = 3.51 g
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<h3>What is decomposition?</h3>

The term decomposition has to do with the breakdown of the given substance into its components. The components of sodium hydrogen carbonate could be identified as water vapor, carbon dioxide gas and sodium carbonate. Among these products that have been listed here, we can see that it is only the sodium carbonate that remains as a solid. The others are gases that move away from the system that is under study.

Now putting down the equation of the reaction, we have;

$2NaHCO_{3} (s) ----- > Na_{2} CO_{3} (s) + CO_{2} (g) + H_{2} O(g)$