Question

26. (a) The given carbon containing compound have 52.17% of Carbon,

13.04% of hydrogen and rest being oxygen. Its molecular mass = 46

gmol−1

. Find its emphirical formula and molecular formula. -4-

(b) Express 27℃ in to Kelvin. -1m​

Answer 1

Answer:

(a) the compond is C2H6O1

(b) 27°c = 300.15 kelvin

method for (a) ->

Question: What is the Empirical formula given the relative number of Atoms, percentage.

Empirical Formula = the smallest whole number ratio of atoms in a chemical formula.

So: Convert with Dimensional Analysis from Percentage mass of elements-> Grams of elements in a Total mass of 100 grams of compound -> Number of moles of each element in a Total mass of 100 grams of compound-> Number of atoms of each element in a Total mass of 100 grams of compound -> Convert to Relative Ratio of atoms by dividing by smallest number or atoms -> Convert to smallest whole number ratio by multiplying by whole numbers if needed.

Percent by mass of a element is a ratio of the element to the total times 100 = (Percent of element ) / ( Percent of a Total compound ) = or (grams of element ) / (in 100 grams of Total compound )

% Element in Compound = 100 * (g Element/g Compound)

100 g Compound * (g Element/g Compound) = g Element

Element Element Precent

C 52.2 % , given

H 13.0 %, given

O 34.8 % = what is left from 100%

Element Precent *100 g Compound *( g Element/ g Compound) = g Element

52.2 % C 52.2 g C

13.0 % H 13.0 g H

34.8 % O 34.8 g O

(g Element )/(molar mass Element(g/mol)) = mol Element

52.2 g C 4.34 mol C

13.0 g H 12.8 mol H

34.8 g O 2.17 mol O

mol Element = Relative # atoms Element

(atoms Element ) /(smallest number atoms) = Ratio of Atoms of Elements in Compound

4.34 atoms C 1.99 atoms C ~ 2 atoms C

12.8 atoms H 5.93 atoms H ~ 6 atoms H

2.17 atoms O 1.00 atoms O 1 atoms O

This should help with most Percent to Empirical Formula problems.

pls mark it as brainliest