Respuesta :
Answer: The molecular formula for the carcinogenic form of asbestos [tex]Mg_3Si_2H_4O_9[/tex]
Explanation:
a) If percentage are given then we are taking total mass is 100 grams.
So, the mass of each element is equal to the percentage given.
Mass of Mg = 26.31 g
Mass of Si= 20.20 g
Mass of H= 1.45 g
Mass of O= (100-(26.31+ 20.20+ 1.45)) = 52.04 g
Step 1 : convert given masses into moles
Moles of Mg=[tex]\frac{\text{ given mass of Mg}}{\text{ molar mass of Mg}}= \frac{26.31g}{24g/mole}=1.10moles[/tex]
Moles of Si=[tex]\frac{\text{ given mass of Si}}{\text{ molar mass of Si}}= \frac{20.20g}{28g/mole}=0.72moles[/tex]
Moles of H=[tex]\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{1.45g}{1g/mole}=1.45moles[/tex]
Moles of O=[tex]\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{52.04g}{16g/mole}=3.25moles[/tex]
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For Mg = [tex]\frac{1.10}{0.72}=1.5[/tex]
For Si =[tex]\frac{0.72}{0.72}=1[/tex]
For H=[tex]\frac{1.45}{0.72}=2[/tex]
For O =[tex]\frac{3.25}{0.72}=4.5[/tex]
The ratio of Mg : Si: H: O = 1.5 : 1 : 2 : 4.5
Converting them into whole numbers :
The ratio of Mg : Si: H: O = 3 : 2 : 4 : 9
Hence the empirical formula is [tex]Mg_3Si_2H_4O_9[/tex]
Empirical mass =[tex]3\times 24+2\times 28+4\times 1+9\times 16=276g[/tex]
Molecular mass = 277 g
[tex]n= \frac{\text {Molecular mass}}{\text {Empirical mass}}=\frac{277}{276}=1[/tex]
Thus molecular formula =[tex]1\times Mg_3Si_2H_4O_9=Mg_3Si_2H_4O_9[/tex]
