To solve for the empirical formula, we write first all the data.
Given:
Compound 1: 76 wt% Ru and 24wt% O
Compound 2: 61.2 wt% Ru and 38.8 wt% O
Required: Empirical Formula of Compound 1
Solution:
Assume total mass of the compound is 100 g
Solving for Compound 1,
76 g Ru x <u>1 mol Ru </u> = 0.75195 mol Ru
101.07 g Ru
24 g O x <u>1 mol O </u> = 1.5 mol O
16 g O
Then, divide each mole with the smallest number of moles calculated
Ru = 0.75195 mol/0.75195 mol = 1
O = 1.5 mol/0.75195 mol = 2
Therefore, the empirical formula for Compound 1 is RuO2.
<em>ANSWER: RuO2</em>
It’s c had this problem last week
The atomic number of an element is characteristic to that element. Atomic number is the number of protons of that element.
Mass number is the sum of protons and neutrons of the element.
since protons and neutrons have a mass of 1 unit each, they together make up the mass of the element.
atomic number - number of protons - 92
mass number - number of neutrons + protons = 234
number of neutrons = mass number - atomic number
neutrons = 234 - 92 = 142
answer is 142
Apply unitary method,
A day = 24 × 60 minutes
So,
Please refer to the attachment above.
Hope this will help you.
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