Chemistry Emprical formula?

Question

A certain anesthetic contains 64.9 percent C, 13.5 percent H, and 21.6 percent O by mass. At 120°C and 750 mmHg, 1.00 L of the gaseous compound weighs 2.30 g. What is the molecular formula of the compound? (Type your answer using the format CH4 for CH4.)

The Emprical formula is C4O10H. At 120°C and 750 mmHg, 1.00 L of the gaseous compound weighs 2.30 g. What is the molecular formula of the compound? I used PV=nRT and got 23.256 for the number of moles. What do i do next to get the molecular formula?

Answer 1

Here is how to do it, you do the math.
part 1 is to calculate the empirical formula.

Assume you have 100 grams, so your percents automatically become grams. Divide each gram mass by the respective molecular mass of each element to get moles.

Convert the moles to whole number ratios, by dividing by the smallest number. Now you have the empirical formula

For the molecular formula you need the molecular mass. Use the ideal gas equation, PV=nRT to get the moles "n". You have the grams of the sample, so that will give you molecular mass. Compare to the empirical formula and you finally get the molecular formula.

Answer 2

i've got in simple terms had a try in this :) a million. ok, first write down the proportion mass of each ingredient. Like this: Sodium - 36.5 Sulphur - 25.5 Oxygen - 39 2. Now divide the mass by using the Relative Atomic Mass. So: Sodium - 36.5/23 = a million.fifty 9 Sulphur - 25.5/32 = 0.seventy 9 Oxygen - 38/sixteen = 2.37 3. Now divide by using the smallest variety: Sodium - a million.fifty 9/0.seventy 9 = 2 Sulphur - 0.seventy 9/0.seventy 9 = a million Oxygen - 2.37/0.seventy 9 = 3 4. So the ratio is two Sodiums, a million Sulphur, 3 Oxygens. 5. The Empirical formulation is Na2SO3. wish this helps!

Answer 3

oh i love these questions!
i hope you don't want us to just do your hw for you :)

to do these types of problems, you want to imagine that you have a 100 g sample of the anesthetic. So that would have 64.9 g of C, 13.5 g H, and 21.6 g O.
divide these numbers respectively by the element's atomic mass,
SO divide 64.9 by 12 blah blah

pick reb's answer it's very well phrased and complete, without actually telling u the answer.
it's your homework!