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AURORKA [14]
3 years ago
11

A 2.52 g sample of a compound containing only carbon, hydrogen, nitrogen, oxygen, and sulfur was burned in excess O2 to yield 3.

52 g of CO2 and 0.839 g of H2O. Another sample of the same compound, of mass 4.14 g, yielded 1.75 g of SO3. A reaction to determine the nitrogen content was carried out on a third sample with a mass of 5.66 g and yielded 1.88 HNO3 (assume that all of the nitrogen in HNO3 came from the compound).
Calculate the empirical formula of the compound.Express your answer as a chemical formula. Enter the elements in the order: C, H, S, N, O.
Chemistry
1 answer:
julsineya [31]3 years ago
5 0

Answer:

The answer to your question is:  C₄H₄NO₂S

Explanation:

Data

2.53 g CvHwNxOySv       ⇒   3.52g CO2  + 0.839g H2O

4.14 g  CvHwNxOySv       ⇒  1.75 g SO3

5.66g  CvHwNxOySv      ⇒    1.88 g HNO3

Empirical formula

For Carbon

MW CO2 = 44g

                         44g of CO2 ------------------- 12 g of C

                          3.52 of CO2 -----------------   x

                            x = (3.52 x 12)/44

                            x = 0.96 g of C

                         12 g  ------------------ 1 mol

                         0.96g ----------------   x

                        x = (0.96 x 1) / 12

                         x = 0.08 mol of C

Percent of Carbon in the sample

                        2.53 g of sample   --------------  100%

                        0.96 g of Carbon   ------------     x

                          x = (0.96 x 100) / 2.53

                          x = 37.94 % of Carbon in the sample

For Hydrogen

MW H2O = 18 g

                        18g of H2O --------------  2 g of H2

                        0.839 g       ------------ -    x

                         x = 0.093 g of H2

                        1 mol of H2 -------------  1 g of H2

                        x                 --------------   0.093 g of H2

                        x = 0.093 mol of H2

Percent of H2 in the sample

                        2.53 g of sample  --------------  100%

                       0.093                    ---------------   x

                        x = 3.68 % of Hydrogen in the sample

For Sulfur

MW of SO3 = 80 g

                       80g of SO3 ----------------  32 g

                       1.75 g          ----------------   x

                        x = 0.7 g of S

                       32g S -----------------   1 mol

                       0.7g -------------------   x

                       x = 0.022 mol of S

Percent of S in the sample

                       4.14 g ------------------  100%

                        0.7 g ------------------   x

                       x = 16.9%

For Nitrogen

MW of HNO3   = 63g

                          63g of HNO3 ---------------- 14 g of N

                           1.88 g           ------------------   x

                          x = 0.42 g of N

                          14 g -----------------   1 mol of N

                          0,42 g of N ------     x

                          x = 0.03 mol of N

Percent of N in the sample

                           5.66 g  ---------------   100%

                           0.42 g -----------------   x

                           x = 7.4%

Percent of Oxygen = 100 - (37.94 + 3.68 + 16.9 + 7.4)

Percent of oxygen = 34.08

Grams of Oxygen                2.53 g of sample ------------- 100%

                                                 x                       ------------   34.08

                                              x = 0.86 g of Oxygen

                                         16 g of O  ----------------- 1 mol

                                         0.86 g of O ---------------  x

                                         x = 0.054 mol of Oxygen

Divide by the lowest # of mol

C =  0.08/ 0.022 = 3.6 ≈ 4

H = 0.093/ 0.022 = 4.2  ≈ 4

N = 0.022/ 0.022 = 1

O = 0.053/ 0.022 = 2.4  ≈ 2

S = 0.03/ 0.022 = 1.4  ≈  1

Empirical formula

                                  C₄H₄NO₂S

                 

                     

                     

You might be interested in
using the equation, C5H12 + 8O2 -> 5CO2 + 6H2O, if 108 g of water are produced, how many grams of oxygen were consumed?
vodomira [7]
Molar mass:

H₂O = 18.0 g/mol

O₂ = 32.0 g/mol

<span>C</span>₅<span>H</span>₁₂<span> + 8 O</span>₂<span> -> 5 CO</span>₂<span> + 6 H</span>₂<span>O
</span>
8 x (32 g ) ------------ 6 x (18 g )
mass O₂ ------------ 108 g H₂O

mass O₂ = 108 x 8 x 32 / 6 x 18

mass O₂ = 27648 / 108

mass O₂ = 256 g

hope this helps!

3 0
3 years ago
What is the ratio of carbon to oxygen in the following balanced equation: 2C + O2–&gt; 2CO
Morgarella [4.7K]

Answer:

The ratio is 2:1, that is 2mol C  / 1mol O2

Explanation:

First check to see if the equation is balanced. It is so now let's find the ratio.

The ratio is the stoichiomeric coefficient of the element or compound.

Carbon : Oxygen

2C : 1 O2   , so 2:1 ..


7 0
3 years ago
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how many kilojoules of energy are necessary to raise the temperature of 3 kilograms of cast iron from 30 degrees Celsius to 120
Scorpion4ik [409]

Specific heat of cast iron is 0.46 kJ / kg K.

Energy change = 0.46 kJ / kg K * 3 kg * (120 - 30) K = 124.2 kJ

5 0
3 years ago
Energy to thermal energy.
sergiy2304 [10]

A hair dryer converts "electrical energy" to "thermal energy".

Option: C

<u>Explanation</u>:

The "electricity" is converted to "heat energy" in a "wire coil". Here electricity process "air" out of "the hair dryer" with the help of forced convection. Most hairdryers use a nichrome wire coil that will not oxidize when heated and allowing it to "blow" the hair dry with high temperature air which speeds up evaporation. Thus "wire heats" the "air faster", in most of the "hair dryers" the "air" is only in "the barrel" for "half of a second".

3 0
3 years ago
A cup of gold colored metal beads was measured to have a mass 425 grams. By water displacement, the volume of the beads was calc
Troyanec [42]

Given parameters;

Mass of gold colored metal beads  = 425g

Volume of water displaced by beads = 48.0cm³

Unknown;

Identity of the metal = ?

Given densities;

Gold: 19.3 g/mL

Copper: 8.86 g/mL

Bronze: 9.87 g/mL

Density is an intensive property of any substance. This implies that we can use the density of any substance to identify it.

Density can be defined as the mass per unit volume of a substance. Every substance has a unique mass per volume.

 Mathematically;

                    Density  = \frac{mass}{volume}

where mass is in kg or g

           volume is in  m³ or cm³

To find the density, we must know the mass and volume.

In this problem, the volume of the gold metal beads is the same as the volume of water displaced. This is a way to measure volume of solids.

Since the volume is given in cm³, and we are comparing with choices that have units in g/mL, we simply convert the volume in cm³ to mL

                   1cm³  = 1mL³

So therefore, volume of gold colored metal is 48mL

Now input the parameters given and solve for the density;

           Density  = \frac{425g}{48mL }    = 8.85g/mL

From the given densities, we clearly see that copper is the metal since they both of similar densities.

4 0
3 years ago
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