Question

you have 18 g of a 12 wt solution of glucose (c6h12o6) in water. calculate the mole percent in the solution ( atomic weights c=12 h=1 o=16 )

Answer 1

Answer on Question #48577, Chemistry, Other

Task:

You have 18 g of a 12% by weight solution of glucose (C₆H₁₂O₆) in water. Calculate the mole percent in the solution (atomic weights C=12, H=1, O=16)?

Answer:

$$\% = \frac{m(C_6H_{12}O_6)}{m(C_6H_{12}O_6) + m(H_2O)} \cdot 100$$ $$m(C_6H_{12}O_6) = \frac{(m(C_6H_{12}O_6) + m(H_2O)) \cdot 100}{100} = \frac{18}{100 \cdot 12} = 2,16\,g$$ $$v = \frac{m}{M}$$ $$M(C_6H_{12}O_6) = 180\,g/mol$$ $$v(C_6H_{12}O_6) = \frac{2,16}{180} = 0,012\,mol$$

Molar percent of each component will be:

$$\%v(C) = \frac{12 \cdot 6}{180} \cdot 100 = 40\%$$ $$\%v(H) = \frac{12}{180} \cdot 100 = 6,7\%$$ $$\%v(O) = \frac{16 \cdot 6}{180} \cdot 100 = 53,3\%$$

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