Writing Empirical Formulas From Percent Composition - Combustion Analysis Practice Problems

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in 12:29, where did you get the 12?

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[@noluthandongwenya8759]

Good evening Sir, can l ask which special integers to take note of when writing the Emperical formula?

[@ainjeffery]

At the very last we find out the molar mass of oxygen(Present in compound) to be something by subtraction the molar masses of C and H from reactants. Where does the other oxygen in air reacting with the compound go?

[@ainjeffery]

+scienceguy I think I solved this problem by watching the video again carefully. If there's a compound of some amount, say, 200 grams and there's C,H and O present then we can break down the 200 into each of those masses. C =x H=y O=z such that x+y+z=200. First we find the masses of C and H from products which turn out to be, say, 150, so that means the sum of x and y has to be 150. Now we just do the arithmetic : x+y+z=200 150+z=200 z=200-150 = 50. So we that means that the 50 mass of oxygen is from the compound and not from the products. *I hope it helped*

[@liamdavis2068]

25:00, wouldn’t the moles of H2O be the same as the moles of Oxygen because for every mole of H2O, isn’t there only 1 oxygen atom?

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that is what i was also thinking

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[@soyadrink9831]

why do we round up for "H" (minute 4:01) it when from 1.99 to 2. However, for 9:51 "H" why doesn't it round up to just 3?

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for the last one what if were not given the mass of the compund 4.5g

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@4:49 How did you get 180 pls

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[@davidvalle3068]

shouldnt you have reported 2 sf for the molar mass of Carbon? at 2.59

[@davidvalle3068]

i used 12.01 for the conversion factor

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[@luluo8056]

This was a bit long. Since we know the total mass of the molecular compound we can multiply the percent composition of each element by 88grams giving us the total amount of grams for each element in the compound. We can then divide that by the mass per mole of each element and figure out how many atoms of each element there are which gives us the complete molecular compound. Then we simplify rations to derive the empirical formula.

[@dostmehmood7599]

Nice

[@ibrahimhamdoun6486]

0.75 g of hydrocarbon compound contains 0.60 g of carbon. (Ar: c=12:0; H=1.0) Which one of the following could be the molecular formula of the hydrocarbon compound? A. C3 H8 B. C2 H4 C. C2 H6 D . C H4 E . C2 H3 Please can you help me solve this question?

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[@catfish1071]

how do you know there is 2 carbons?

[@catfish1071]

sorry meant hydrogen

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tell me the empirical formula of urea

[@debendramishra6496]

3:13 it's actually 16g O not 16g H.

[@christinemusonda5342]

Where is the 216coming from

[@alsh7107]

The light in the dark.

[@carefull8365]

How much do u round it

[@ariannaethel8991]

Im still confused with how and why you multiplied 3 with the empirical formula with decimal.

[@michelleceniceros4841]

me too, I thought you would just round 2.67 to 3?

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