## Chemistry: The Central Science (13th Edition)

1. Cover the mass of $C_6H_{12}O_6$ to nº of moles: $mm (C_6H_{12}O_6) = 12*6 + 1*12 + 16*6 = 72 + 12 + 96 = 180$ $n(moles) = \frac{mass(g)}{mm}$ $n(moles) = \frac{16}{180} \approx 0.0888$ 2. Now, convert that number to kJ: $0.0888mol(C_6H_{12}O_6) \times \frac{2812}{1mol(C_6H_{12}O_6)}$ $249.7kJ$ 3. Finally, convert that number to calories: $249.7kJ \times \frac{1 Cal}{4.184kJ}$ $\approx 59.68$ $Cal$