Answer : The sample is, CO (carbon monoxide)
Solution : Given,
Mass of C = 24.50 g
Mass of O = 32.59 g
Molar mass of C = 12 g/mole
Molar mass of O = 16 g/mole
First convert given masses into moles.
Moles of C = [tex]\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{24.50g}{12g/mole}=2.04moles[/tex]
Moles of O = [tex]\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{32.59g}{16g/mole}=2.04moles[/tex]
Now for the mole ratio, divide each value of moles by the smallest number of moles calculated.
For C = [tex]\frac{2.04}{2.04}=1[/tex]
For O = [tex]\frac{2.04}{2.04}=1[/tex]
The ratio of C : O = 1 : 1
The mole ratio of the element is represented by subscripts in empirical formula.
The Empirical formula = [tex]C_1O_1[/tex] or, [tex]CO[/tex]
Therefore, the sample is, CO (carbon monoxide)
The sample is carbon monoxide, CO
To obtain the name of the sample, we shall determine the empirical formula of the compound. This can be obtained as follow:
Carbon (C) = 24.50 g
Oxygen (O) = 32.59 g
C = 24.50 / 12 = 2.04
O = 32.59 / 16 = 2.04
C = 2.04 / 2.04 = 1
O = 2.04 / 2.04 = 1
Thus, the empirical formula of the compound is CO.
The name of the CO is carbon monoxide.
Therefore, we can conclude that the sample is carbon monoxide, CO.
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