What answer should be reported, with the correct number of significant figures,
ID: 1001912 • Letter: W
Question
What answer should be reported, with the correct number of significant figures, for the following calculation? (249.362 +41)/63.498 = 2. In 0.50 mole of methyl formate, HCOOCH_3, there are a. C atoms b. H c. Total # of atoms. d. What is the mass of 0.5 mole of methyl formate? e. What is the mass of 2.4times10^23 molecules of methyl formate? Determine the molecular formula of a compound that is 49.48% carbon, 5.19% hydrogen, 28.85% nitrogen, and 16.48% oxygen. The molecular weight is 194.19 g/mol.Explanation / Answer
Solutions:- (1) here we have two maths operations, addition and division. First we would solve for the parenthesis. There is addition of two numbers in the parenthesis so let's do the addition and as per the rules for sig figs, " In addition we go with least decimal places."
249.362 has two decimal places wheer as 41 has no decimal place so the answer for addition step would not have any decimal place.
249.362 + 41 = 290 (no decimal place)
in the next step we would divide 290 by 63.498 and as per the rules for sig figs, " in multiply and division we go for least number of sig figs. 290 has 2 sig figs where as 63.498 has 5 sig figs. Here, least sig figs are 2 so we would round of the answer to two sig figs.
290/63.498 = 4.567 = 4.6 (Answer with two sig figs)
(2) HCOOCH3 has 4 hydrogen, 2 carbon and 2 oxygen.
It means 1 mole of HCOOCH3 has 4 moles of H, 2 moles of C and 2 moles of O.
we also know that 1 mol of anything equals to 6.022 x 1023(Avogadro number). this means 1 mol of each mol of atoms would be equal to 6.022 x 1023 atoms.
(a) 0.50 mol HCOOCH3 x (2 mol C/1mol HCOOCH3) x (6.022 x 1023 C atoms/1mol C) = 6.022 x 1023 C atoms
(b) 0.50 mol HCOOCH3 x (4 mol H/1mol HCOOCH3) x (6.022 x 1023 H atoms/1mol H) = 1.204 x 1024 H atoms
(c) For total atoms we need to calculate the atoms of oxygen also.
0.50 mol HCOOCH3 x (2 mol O/1mol HCOOCH3) x (6.022 x 1023 O atoms/1mol O) = 6.022 x 1023 O atoms
so, total atoms = 6.022 x 1023 + 6.022 x 1023 + 1.204 x 1024
= 6.022 x 1023 + 6.022 x 1023 + 12.04 x 1023
= 1023(6.022 + 6.022 + 12.04) = 24.084 x 1023 = 2.4084 x 1024 atoms
(d) to convert the moles into grams we multiply the moles by molar mass. Molar mass is the sum of atomic masses of all the atoms present in the compound.
molar mass of methyl formate = 2x12 + 4x1 + 2x16 = 24+4+32 = 60 g/mol
0.5 mol x (60g/1mol) = 30 g
(e) First we would convert the molecules into moles on dividing the molecules by Avogadro number and then follow the same step that we did for (d).
2.4 x 1023 molecules x (1mol/6.022 x 1023 molecules) x (60g/1mol) = 23.91 g and it could be round to 24 g.
(3) Firtst of all we find out the empirical formula for the given percentages. Sum of the percentages is equal to 100 so their percentages are taken as their masses and divide these by their corresponding atomic masses to get their moles.
49.48 g of C x (1mol of C/12 g of C) = 4.12 mol of C
5.19 g of H x (1mol of H/1 g of H) = 5.19 mol of H
28.85 g of N x (1mol of N/14 g of N) = 2.06 mol of N
16.48 g of O x (1mol of O/16 g of O) = 1.03 mol of O
Now we divide all of the by the least one to get the mol ratio and least one is moles of O. So, divide all by 1.03
C = 4.12/1.03 = 4
H = 5.19/1.03 = 5
N = 2.06/1.03 = 2
O = 1.03/1.03 = 1
So, the empirical formula is C4H5N2O
Empirical formula mass = 4x12 + 5x1 + 2x14 + 1x16 = 48+5+28+16 = 97 g/mol
Molar mass is given as 194.19g/mol
we will divide the molar mass by empirical formula mass to calculate the empirical formula units as molecular formula contains n number of empirical formual units where n is an integer like 1,2,3 and so on.
n = 194.19/97 = 2
so, molecular formula contains two empirical formula units ( C4H5N2O)
so, molecular formula = ( C4H5N2O)2 = C8H10N4O2
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