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A 230 g sample of a compound contains 136.6 g of carbon. 26.4 g of hydrogen, and 31.8 g of nitrogen. The rest is oxygen. What is the mass percent of oxygen in the compound?
The mass of oxygen in the $230$ $g$ sample is simply that mass that is not $C$, $H$, or $N$. So I simply subtract from the starting mass and the...
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A compound containing only carbon and hydrogen has a carbon-to-hydrogen mass ratio of 11.89. Which carbon-to-hydrogen mass ratio is possible for another compound composed only of carbon and hydrogen?
Trying to present a possible Answer We know that the molar mass of carbon and hydrogen are $C=12"g/mol" and H = 1"g/mol"$ When Carbon Hydrogen mass ratio is 11.89 The...
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Elemental analysis of a compound showed that it consisted of 81.82% carbon and 18.18% hydrogen by mass, how many hydrogen atoms appear in the empirical formula of the compound?
For this kind of problem, we have to assume that you have 100g unknown sample (since the percentages add up to 100%). Thus, masses are C = 81.82 g; H...
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Elemental analysis of a compound showed that it consisted of 81.82% carbon and 18.18% hydrogen by mass. How many hydrogen atoms appear in the empirical formula of the compound?
Since the percentages add up to 100, we can assume that we have a 100.0 g sample of the compound, and the percentages become grams. Determine the Moles of Each...
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A compound that contains only nitrogen and oxygen is 30.4% N by mass; the molar mass of the compound is 92 g/mol. What is the empirical formula and the molecular formula of the compound?
In $100* g$ of this there are $30.4* g$ $N$, and the balance $O$. We divide thru by the atomic masses in order to approach the empirical formula: $(30.4*g)/(14.01*g*mol^-1)$ $=$...
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A 5.325 g sample of methyl benzoate, a compound in perfumes, was found to contain 3.758 g of carbon, 0.316 g of hydrogen, and 1.251 g of oxygen. What is the empirical formula? If its molar mass is about 136 g/mol, what is its molecular formula?
I propose an interesting alternative to the classic approach of finding the empirical formula first, then using the molar mass to find the molecular formula. More specifically, we will find...
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Combustion of a 0.9827 g sample of a compound containing only carbon, hydrogen, and oxygen produced 1.900 g of $CO_2$ and 1.070 g of $H_2O$. What is the empirical formula of the compound?
Here is the equation with masses: $"0.9827 g C"_a"H"_b"O"_c + "(1.900 g + 1.070 g - 0.9827 g) O"_2 rarr "1.070 g H"_2"O" + "1.900 g CO"_2$ Use 32.00 g/mol...
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Another compound found on Mars contains iron and oxygen. The compound contains 70% by mass of iron and 30% by mass of oxygen. How do you calculate the empirical formula of this compound?
Your ultimate goal here is to figure out the smallest whole number ratio that exists between iron and oxygen in this unknown compound, i.e. the empirical formula of the compound....
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Combustion analysis of a 12.01 g sample of tartaric acts - which contains only carbon, hydrogen, and oxygen - produced 14.08 g CO2 and 4.32 g H2O. What is the empirical formula for tartaric acid?
We address the molar quantities of carbon, and hydrogen .... convert these into masses, and then address the empirical formula. $"Moles of carbon dioxide"=(14.01*g)/(44.01*g*mol^-1)=0.3183*mol$ And thus mass of carbon...$-=0.3183*molxx12.011*g*mol^-1=3.824*g$ $"Moles...
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)A compound that contains just C, H and N is burned in oxygen. When 5.000 g of the compound is burned in excess oxygen, 6.019 g of $CO_2$, 4.313 g of $H_2O$, and 9.438g of $NO_2$ are the only products. What is the empirical formula of the compound?
We must calculate the masses of $"C, H"$, and $"N"$ from the masses of carbon dioxide, water, and nitrogen dioxide. $"Mass of C" = 6.019 color(red)(cancel(color(black)("g CO"_2))) × "12.01 g...
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