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Romashka-Z-Leto [24]
2 years ago
15

Caffeine, a molecule found in coffee, tea, and certain soft drinks, contains C, H, O, and N. Combustion of 10.0 g of caffeine pr

oduces 18.13 g of CO₂, 4.639 g of H₂O, and 2.885 g of N₂. Determine the molar mass of the compound if it is between 150 and 210 g/mol.
Chemistry
1 answer:
denis-greek [22]2 years ago
8 0

Answer:

194 g/mol.

Explanation:

Hello,

In this case, one first must compute the mass of each element as shown below:

C=18.13gCO_2*\frac{12gC}{44gCO_2} =4.945gC\\H=4.639gH_2O*\frac{2.016gH}{18.0152gH_2O}=0.519gH\\N=2.885gN_2\\O=10.0g-4.945g-0.519g-2.885g=1.651gO

Next, the corresponding moles:

C=4.945gC*\frac{1molC}{12gC}=0.412mol\\H=0.519gH*\frac{1molH}{1gH}=0.519mol\\N=2.885gN*\frac{1molN}{14gN}=0.206molN\\O=1.648gO*\frac{1molO}{16gO} =0.103molO

Then, each element's subscripts is found to be:

C=\frac{0.412}{0.103}=4\\H=\frac{0.519}{0.103}=5\\N=\frac{0.206}{0.103} =2\\O=\frac{0.103}{0.103}=1

Therefore, the empirical formula is:

C_4H_5N_2O

Nonetheless, it has a molar mass of 97bg/mol, thereby, by multiplying such formula by 2 one gets:

C_8H_10N_4O_2

Which has a molar mass of 194 g/mol being correctly contained in the given interval.

Best regards.

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Answer:

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Explanation:

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1 year ago
How many sodium atoms are in 0.1310 g of sodium
Lana71 [14]
To convert grams to atoms, we first need to find the moles and then multiply by Avogadro's constant.

0.1310 g * (1 mol/22.99 g) * (6.022*10^23 atom/1 mol) = 3.431 *10^21 atoms.
3 0
2 years ago
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58.5g of sodium iodide are dissolved in 0.5dm3. what concentration would this be?
Bezzdna [24]
percent by mass=(mass of solute/ mass of solution)*100 %

mass of solute=58.5 g

density (H₂O)=1 g/cm³*(1000 cm³/1dm³)=1000 g/dm³

mass of solvent (H₂O)=0.5 dm³ * (1000 g/dm³)=500 g
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mass of solution=58.5 g+500 g=558.5 g

% mass=(58.5 g/558.5 g) * 100%=10.47% of Na.

solution:  10.47% of Na.
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How many moles of tungsten (W,183.85 g/lol are in 415 grams of tungsten?
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Given mass of tungsten, W = 415 g

Molar mass of tungsten, W = 183.85 g/mol

Calculating moles of tungsten from mass and molar mass:

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7 0
1 year ago
In this experiment, 0.170 g of caffeine is dissolved in 10.0 ml of water. the caffeine is extracted from the aqueous solution th
zmey [24]

solution:

Weight of caffeine is W = 0.170 gm.

Volume of water is V= 10 ml

Volume of methylene chloride which extracted caffeine is v= 5ml

No of portions n=3

Distribution co-efficient= 4.6

Total amount of caffeine that can be unextracted is given by

w_{n}=w\times[\frac{k_{Dx}v}{k_{Dx}v+v}]^n\\w_{3}=0.170[\frac{4.6\times10}{(4.6\times10+5)}]^3\\=0.170[\frac{46}{46+5}]^3\\=0.170[\frac{46}{51}]^3\\=0.170[\frac{97336}{132651}]\\=0.170\times0.734=0.125gms

amount of caffeine un extracted is 0.125gms

amount of caffeine extracted=0.170-0.125

                                                       =0.045 gms


6 0
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