A solution contains 10.05 g of unknown compound dissolved
in 50....

Question

A solution contains 10.05 g of unknown compound dissolved 
in 50.0 mL of water. (Assume a density of 1.00 g>mL for water.) 
The freezing point of the solution is -3.16 °C. The mass percent 
composition of the compound is 60.97% C, 11.94% H, and the 
rest is O. What is the molecular formula of the compound?

Accepted Answer

Hi everyone here we have a question telling us that an unknown compound is made up of only carbon, hydrogen and oxygen with a percent composition of 62.4% carbon and 10.4 point 1% hydrogen. A massive 14.3 g of this compound was dissolved in 70.5 millions of water D equals 1.0 g per milliliter. The freezing point of the solution is negative 3.05°C determine the molecular formula of the compound, so we know that the change in freezing point equals the freezing point of our peer solvent minus the freezing point of our solution. The change in our freezing point also equals r rant. Hoff factor times are freezing point constant times morality. Our freezing point constant for water Equals 1.86° C per morality. And the freezing point for our water is 0°C. We're going to assume that I equals one that is a Covalin molecule. So now we're going to calculate morality. So the change in freezing point equals the freezing point of our solvent minus the freezing point of our solution. As we said earlier, Which means 0°C -3.05°C Equals 3.05°C. So our morality Equals 3.05°C times one Times 1.86°C per morality Equals 1. morality, Which equals 1. 398 moles per kilogram. So now to calculate our molar mass, the moles of our compound Equals 75 ml times one g per millimeter times one kg Per 10 to the 3rd g Times 1.639, 8 moles per kilogram. And our middle leaders are going to cancel out. Our g are going to cancel out. Our kg are going to cancel out leaving us with 0. moles. So our molar mass Equals 14.3 g, divided by 0.123 moles. That equals 116.26 g per mole. So now we have .04 carbon and 10. hydrogen And 27.55% oxygen. So -62.04% carbon -10.4, Equals 27.55%. That's how we got the 27.55% oxygen. And now we're going to calculate our empirical formula. We're gonna assume that we have 100g of the compound. So the moles of carbon Equals 62. grams carbon times one mole carbon Over 12.01 g carbon Equals 5. moles. Our moles of hydrogen Equals 10.41g times one mole hydrogen Over 1.008g hydrogen. And our g are canceling out giving us 10.3274 moles. Our moles of oxygen Equals 27.55g of oxygen times one mole Divided by 16 g of oxygen. Our g are canceling out, giving us 1.7219 moles. And then we're going to divide each of these by the smallest number of moles, so 1.7-9. And lastly Our 5.1657. So our empirical formula equals C three H 60. Our E. F. Mass Equals 12.01 g per mole times three plus 1.8 g Permal Times six plus 16.00 g. Permal That equals 58.078 g per mole. And now to calculate our ratio, we have 116. g Permal, Divided by 58.078 g per mole that equals two. So we need to multiply everything in our empirical formula two, which gives us C six H 12 02. And that is our final answer. Thank you for watching. Bye.

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