Answer:
C2H2O4
Explanation:
To get the molecular formula, we first get the empirical formula. This can be done by dividing the percentage compositions by the atomic masses. The percentage compositions are shown as follows :
C = 26.86%
H = 2.239%
O = 100 - ( 26.86 + 2.239) = 70.901%
We then proceed to divide by their atomic masses. Atomic mass of carbon is 12 a.m.u , H = 1 a.m.u , O = 16 a.m.u
The division is as follows:
C = 26.86/12 = 2.2383
H = 2.239/1 = 2.239
O = 70.901/16 = 4.4313
We now divide each by the smallest number I.e 2.2383
C = 2.2383/2.2383 = 1
H = 2.239/2.2383 = 1
O = 4.4313/2.2383 = 1.98 = 2
Thus, the empirical formula is CHO2.
To get the molecular formula, we use the molar mass .
(CHO2)n = 90
We add the atomic masses multiplied by n.
(12 + 1 + 2(16))n = 90
45n = 90
n = 90/45 = 2.
Thus , the molecular formula is C2H2O4
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Answer:
the conversion factor is f= 6 mol of glucose/ mol of CO2
Explanation:
First we need to balance the equation:
C6H12O6(s) + O2(g) → CO2(g) + H2O(l) (unbalanced)
C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l) (balanced)
the conversion factor that allows to calculate the number of moles of CO2 based on moles of glucose is:
f = stoichiometric coefficient of CO2 in balanced reaction / stoichiometric coefficient of glucose in balanced reaction
f = 6 moles of CO2 / 1 mol of glucose = 6 mol of glucose/ mol of CO2
f = 6 mol of CO2/ mol of glucose
for example, for 2 moles of glucose the number of moles of CO2 produced are
n CO2 = f * n gluc = 6 moles of CO2/mol of glucose * 2 moles of glucose= 12 moles of CO2
Answer:
1.13 moles Au
Explanation:
Moles Au = 6.80x10²³atoms / 6.023x10²³atoms/mole = 1.13 moles Au
