9.03 X [tex]10^{22}[/tex] molecules of butane is present in 9.213 grams of the sample.
1.5 atoms of hydrogen are present in the sample 9.123 grams of butane.
Explanation:
Data given:
mass of butane C4H10= 9.123 grams
atomic mass of butane = 58.12 grams/mole
number of molecules =?
number of hydrogen atoms in the sample=?
number of moles is calculated as :
number of moles = [tex]\frac{mass}{atomic mass of 1 mole}[/tex]
putting the values in the above equation:
number of moles of butane = [tex]\frac{9.213}{58.12}[/tex]
= 0.15 moles
number of molecules = number of moles x 6.02 x [tex]10^{23}[/tex] (Avagadro number)
Putting the values in above formula:
number of molecules of Butane = 0.15 X 6.023 X [tex]10^{23}[/tex]
= 9.03 X [tex]10^{22}[/tex] molecules
Number of hydrogen atoms:
1 mole of C4H10 contains 10 atoms of H
0.15 moles of C4H10 will have x moles
[tex]\frac{10}{1}[/tex] = [tex]\frac{x}{0.15}[/tex]
x = 0.15 atoms of hydrogen
There are 9.6 × 10^23 atoms of hydrogen in the sample.
Now we must first obtain the number of moles that are contained in 9.213 g sample of butane.
Molar mass of butane = 4(12) + 10(1) = 48 + 10 = 58 g/mol
Number of moles of butane = 9.213 g/58 g/mol = 0.16 moles
Number of hydrogen atoms in 9.213 g sample of this sample can now be obtained by 0.16 × 6.02 × 10^23 × 10 = 9.6 × 10^23 atoms of hydrogen.
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