Answer:
[tex]C_3H_7O_3[/tex]
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to infer that the empirical formula of mannitol contains carbon, hydrogen and oxygen, so that the first step is to calculate the moles of C and H contained in the CO2 and H2O, respectively, as the only sources of these two elements in the formula:
[tex]n_C=1.993gCO_2*\frac{1molCO_2}{44.01gCO_2}*\frac{1molC}{1molCO_2} =0.0453molC\\\\n_H=0.9519gH_2O*\frac{1molH_2O}{18.02gH_2O}*\frac{2molH}{1molH_2O} =0.106molH[/tex]
Next, we calculate the grams and moles of O by subtracting the mass of C and H from the mass of the sample:
[tex]m_O=1.375g-0.0453molC*\frac{12gC}{1molC}-0.106molH*\frac{1.01gH}{1molH}=0.724gO\\\\n_O=0.724gO*\frac{1molO}{16.0gO} =0.0453molO[/tex]
Finally, we divide the moles of C, H and O by 0.0453 as the fewest moles of both C and O to find the mole ratios in the formula:
[tex]C:\frac{0.0453mol}{0.0453mol} =1\\\\H:\frac{0.106mol}{0.0453mol} =2.34\\\\O:\frac{0.0453mol}{0.0453mol} =1[/tex]
To get:
[tex]CH_{2.34}O[/tex]
Which must be multiplied by 3 to get whole numbers for all the subscripts, and therefore obtain:
[tex]C_3H_7O_3[/tex]
Regards!
