The grams of 22.9 % sugar solution that contain 68.5 g of sugar is 299.13 g of solution
<u><em>calculation</em></u>
22.9% means that there are 22.9 g of sugar in 100 g of solution.
what about 68.5 g of sugar
- <em>by cross multiplication</em>
=[(68.5 g sugar x 100 g solution) /22.9 g sugar] =299.13 g of solution
Nb; <em>g sugar cancel each other</em>
<span>Mass Number = (Atomic Number) + (Number of Neutrons) so you solve for the Number of Neutrons and you get:
Number of Neutrons = (Mass number) - (Atomic Number)
Mass Number equals protons plus neutrons, round atomic weight to nearest whole number
Atomic Number equals number of Protons</span>
The empirical formula is SCl_2.
The <em>empirical formula</em> (EF) is the simplest whole-number ratio of atoms in a compound.
The ratio of atoms is the same as the ratio of moles.
So, our job is to calculate the <em>molar ratio </em>of S to Cl.
Assume that you have 100 g of sample.
Then it contains 31.14 g S and 68.86 g Cl.
<em>Step</em> 1. Calculate the <em>moles of each element</em>
Moles of S = 31.14 g S × (1 mol S/(32.06 g S) = 0.971 30 mol S
Moles of Cl = 68.86 g Cl × (1 mol Cl/35.45 g Cl) = 1.9425 mol Cl
<em>Step 2</em>. Calculate the <em>molar ratio</em> of each element
Divide each number by the smallest number of moles and round off to an integer
S:Cl = 0.971 30: 1.9425 = 1:1.9998 ≈ 1:2
<em>Step 3</em>: Write the <em>empirical formula</em>
EF = SCl_2
