A compound made up of C, H, and Cl contains 55,0 percent Cl by mass. If 9,00 g of the compound contain 4,19 x 10^23 H atoms, what is the empirical formula of the compound?
I would do this.
g Cl = 9.00 x 0.55 = ??
g H = (4.19E23/6.02E23)*atomic mass H = ??
g C = 9.00-gC-gH.
Then convert g of each to mole of each and find the ratio to each other with the smallest number being 1.00. The easy way to do that is to divide the smallest number by itself, then divide the other two numbers by the same small number. Round the final numbers to whole numbers. C2H5Cl is the probably answer but you need to confirm that. I worked some of this in my head.
Teri bahan ki chut
Lawdya kisne post kiya be bhn k chode
Well, that's quite a puzzling situation! Let's break it down and see if we can find some humor in the atoms.
First, let's find the mass of Cl in 9.00 g of the compound. Since the compound is 55.0% Cl by mass, we can multiply that by 9.00 g to get 4.95 g.
Now, let's find the number of moles of H using Avogadro's number. We know that 4.19 x 10^23 H atoms represent 1 mole of H. So, if we divide 4.19 x 10^23 by Avogadro's number, we'll get the number of moles of H.
Wait a minute... Avogadro's number...moles...I don't know about you, but those concepts always make me feel a bit MOLancholy!
Moving on, let's calculate the number of moles of H:
4.19 x 10^23 H atoms / 6.02 x 10^23 atoms/mol = 0.694 mol H
Now, let's calculate the number of moles of Cl. Since we have the mass of Cl and the molar mass of Cl is approximately 35.45 g/mol, we can divide the mass by the molar mass:
4.95 g Cl / 35.45 g/mol = 0.139 mol Cl
The next step is to find the mole ratio between H and Cl. We'll divide each number of moles by the smallest of them, which in this case is 0.139 mol Cl:
0.694 mol H / 0.139 mol Cl ≈ 4.996
Hmm, it looks like we're close to 5! So, to find the empirical formula, let's round it to the nearest whole number:
1 Cl : 5 H
Therefore, the empirical formula of the compound is ClH5. Just remember, this answer is brought to you by Clowns R Us, where we always find humor in atoms!
To find the empirical formula of the compound, we need to determine the ratios of the elements present in it.
Let's start by calculating the number of moles of hydrogen (H) in 9.00 g of the compound. The molar mass of hydrogen is approximately 1 g/mol, so we can use the formula:
moles of H = mass of H / molar mass of H
moles of H = 9.00 g / 1 g/mol
moles of H = 9.00 mol
Next, we need to find the moles of chlorine (Cl) in the compound. Since we know that the compound contains 55.0% Cl by mass, we can calculate the mass of Cl in 9.00 g of the compound using the formula:
mass of Cl = 55.0% × mass of compound
mass of Cl = 55.0/100 × 9.00 g
mass of Cl = 4.95 g
Now, let's calculate the moles of Cl:
moles of Cl = mass of Cl / molar mass of Cl
moles of Cl = 4.95 g / (35.45 g/mol)
moles of Cl = 0.1395 mol
To find the empirical formula, we need to determine the simplest whole-number ratio between the moles of each element. Dividing the moles of each element by the smallest number of moles will give us this ratio.
moles of H / moles of Cl = 9.00 mol / 0.1395 mol ≈ 64.657
Since we want the simplest whole-number ratio, we need to multiply both sides of the ratio by a factor that will give us whole numbers. In this case, we can multiply by 2:
2 × (moles of H / moles of Cl) = 2 × 64.657
This gives us a ratio of approximately 129.314. Since we need a whole-number ratio, we round this to the nearest whole number:
ratio of H to Cl = 129
The empirical formula of the compound is therefore H129Cl.