CH3 is the empirical formula for the compound.
A sample of a compound is determined to have 1.17g of Carbon and 0.287 g of hydrogen.
The number of atom or moles in the compound is
1.17 g C X 1 mol of C / 12.011 g C = 0.097411 mol of C.
0.287 g H x 1 mol of H / 1 g H = 0.28474 mol H.
This compound contains 0.097411 mol of carbon and 0.28474 mol of Hydrogen.
So we can represent the compound with the formula C0.974H0.284.
Subscripts in formulas can be made into whole numbers by multiplying the smaller subscript by the larger subscript.
we can divide 0.284 by 0.0974.
0.284 / 0.0974 = 3.
So here, Carbon is one and hydrogen is 3.
We can write the above formula as a CH3.
Hence the empirical formula for the sample compound is CH3.
For a detailed study of the empirical formula refer given link brainly.com/question/13058832.
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Answer:
I would recommend them becoming an Analytical Chemist because Analytical Chemists examine and identify various elements and compounds to find out the composition, structure, and nature of substances and they determine the concentration of chemical pollutants in soil, water,and air. I would recommend taking inorganic, organic, analytical, and physical chemistry as well as computer science, physics, and environmental science
<u>8:4 and 4:8</u> both have CCP structure............................... ᐛ
Answer:
Cost to supply enough vanillin is 
Explanation:
Threshold limit of vanillin in air is 
per litre means there should be of vanillin in 1L of air to detect aroma of vanillin.
So, 
So amount of vanillin should be present to detect = 
As cost of 50 g vanillin is 
therefore cost of vanillin = 
2(units), Force = mass * acceleration, so mass = force/acceleration, which is 2
