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Misha Larkins [42]
3 years ago
14

what is the empirical formula for a substance if a 1.000 g sample of it contains 0.262 grams of nitrogen, 0.075 grams of hydroge

n, and 0.663 grams of chlorine
Chemistry
1 answer:
Sonbull [250]3 years ago
4 0
Here are the steps for finding an emprical formula.

To begin, we need to change the grams to a percentage. We can simply take the number and call it our percentage. (i.e 0.262%)

Then, we have a simply stoichiometry step. We put our starting number over 1, as usual. The next "Step" is putting in the molar mass of that particular element on the bottom, with one on top. It should look like this: 

0.262% N          1
-------------- x ----------------      
       1              14.007g N    

We would then divide those two numbers to get our answer. Repeat for the other elements included, until we have all of them done.

After that, we need to find the smallest number that was the answer to one of the equations. For example, if the product of Hydrogen was the smallest formed, we would locate that number. We then divide ALL of our answers by that smallest number.

Remember, whenever doing empirical formulas, significant figures matter.

Once we have our smallest numbers, we need to round our numbers to the nearest whole number. HOWEVER, to do this we need to multiply all of the numbers by the SAME whole number.

For example, if one of our numbers was 1.33, we would multiply by 3 to get a whole number. However, we would also need to multiply the rest of our products by the same number. If we do not get close to whole numbers (you can round), we need to try with multiplying a different number.

Then, we need to take those whole numbers we got for each number, and plug in to make a compound. For example, if our products were 6 for nitrogen, 3 for hydrogen, and 2 for chlorine, our final empirical formula would be N6H3Cl2.

Hope I could help! If it was a bit wordy please let me know and I'd be more than willing to clarify.   
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Can anybody check my answer?
anzhelika [568]

Answer:

\boxed{\text{25. 20 L; 26. 49 K}}

Explanation:

25. Boyle's Law

The temperature and amount of gas are constant, so we can use Boyle’s Law.

p_{1}V_{1} = p_{2}V_{2}

Data:

\begin{array}{rcrrcl}p_{1}& =& \text{100 kPa}\qquad & V_{1} &= & \text{10.00 L} \\p_{2}& =& \text{50 kPa}\qquad & V_{2} &= & ?\\\end{array}

Calculations:

\begin{array}{rcl}100 \times 10.00 & =& 50V_{2}\\1000 & = & 50V_{2}\\V_{2} & = &\textbf{20 L}\\\end{array}\\\text{The new volume will be } \boxed{\textbf{20 L}}

26. Ideal Gas Law

We have p, V and n, so we can use the Ideal Gas Law to calculate the volume.

pV = nRT

Data:  

p = 101.3 kPa

V = 20 L

n = 5 mol

R = 8.314 kPa·L·K⁻¹mol⁻¹

Calculation:

101.3 × 20 = 5 ×  8.314 × T

2026 = 41.57T

T = \dfrac{2026}{41.57} = \textbf{49 K}\\\\\text{The Kelvin temperature is }\boxed{\textbf{49 K}}

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4 years ago
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