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3 years ago

How many grams of hydrogen are in 92 g of ch4o?

Chemistry

2 answers:

IRISSAK [1]3 years ago

7 0

The answer to this question would be: 11.5 grams

The molecular mass of hydrogen is 1, oxygen is 16 and carbon is 12. Then the molecular mass of CH4O would be: 12 + 4(1) + 16= 32.
To find the hydrogen mass, you just need to divide the hydrogen with total mass, resulting in the mass ratio of hydrogen to the compound. Then the amount of hydrogen inside 92 gram of CH4O are: 92grams * (4/32)= 11.5 grams

Hoochie [10]3 years ago

5 0

The mass of Hydrogen in 92 g of CH4O is 11.5 grams

<h3><em>Further explanation </em></h3>

Proust stated the Comparative Law that compounds are formed from elements with the same Mass Comparison so that the compound has a fixed composition of elements

In the same compound, although from different sources and formed by different processes, it will still have the same composition/comparison

With this law, we can calculate how many grams an element is needed to make a compound with a certain mass, as desired

The mass of compounding compounds can also be calculated.

For example, the mass ratio of oxygen (O) and hydrogen (H) elements in H₂O remains at 8: 1. So if you know the amount of mass of oxygen you can know the amount of hydrogen and water mass.

Or the comparison of CO₂ with C: O = 3: 8

Mass A in AxBy = (x x Ar A x mass AₓBy) / Mr AₓBy

Mass B in AₓBy = (y x Ar B x mass AₓBy) / Mr AₓBy

Note that x, y are the numbers of A and B in AₓBₐ

In CH₄O compounds (Methanol, with relative molecular mass = 32) has a mass of 92 grams and there are 4 Hydrogen atoms, so the mass is:

$mass~H~=~\frac{(4\times relative~atomic~mass~H}{relative~molecular~mass ~CH_4O}~\times~mass~CH_4O$

$mass~H~=~\frac{4\times1}{32} \times~92$

mass H = 11.5 grams

<h3><em>Learn more </em></h3>

A sample of H₂SO₄

brainly.com/question/4310288

mole Caffeine

brainly.com/question/9505286

Given a compound with 169 amu

brainly.com/question/5303004

Keywords: Proust, the Comparative Law, CH₄O

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3 years ago

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a. 381.27 m/s

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<h3>Further explanation</h3>

Given

T = 100 + 273 = 373 K

Required

a. the gas speedi

b. The rate of effusion comparison

Solution

Average velocities of gases can be expressed as root-mean-square averages. (V rms)

$\large {\boxed {\bold {v_ {rms} = \sqrt {\dfrac {3RT} {Mm}}}}$

R = gas constant, T = temperature, Mm = molar mass of the gas particles

From the question

R = 8,314 J / mol K

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Mm = molar mass, kg / mol

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