1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Gnesinka [82]
3 years ago
12

Can someone please explain how to find an emperical formula given mass not percentage?​

Chemistry
1 answer:
Firdavs [7]3 years ago
8 0

Answer:

1. Get the mass of each element by assuming a certain overall mass for the sample (100 g is a good mass to assume when working with percentages). ...

2. Convert the mass of each element to moles. ...

3. Find the ratio of the moles of each element. ..

4. Use the mole ratio to write the empirical fomula

You might be interested in
G x mol/g x atoms/mol= <br><br> what are the units remaining after conversion
antiseptic1488 [7]

Answer:

The "unit" remaining after conversion is atoms.

This is a way to calculate the number of atoms, in an specific mass.

Explanation:

When you have a mass, multiplying the molar mass, "grams" are cancelled, so you finally obtained, number of moles.

Moles . atoms/mol, cancel the word mol so you will find out a number of atoms, present at the mass from, the begining.

For example, how many atoms are in 10 g of NaCl

Molar mass NaCl = Na mass + Cl mass

23 + 35.45 = 58.45 g/m

10 g . 1 / 58.45 mol/g  . NA/ 1 mol = 3.42x10²⁰ atoms

NA = Avogadro number

3 0
4 years ago
Calculate the empirical formula of a compound that is 42.9% Carbon and 57.1% Oxygen.
Marrrta [24]

Answer:

CO.

Explanation:

Assuming the given percentages are by mass, we can solve this problem via imagining we have <em>100 g of the compound</em>, if that were the case we would have:

  • 42.9 g of C
  • 57.1 g of O

Now we <u>convert those masses into moles</u>, using the<em> elements' respective molar masses</em>:

  • 42.9 g of C ÷ 12 g/mol =  3.57 mol C
  • 57.1 g of O ÷ 16 g/mol =  3.58 mol O

As the number of C moles and O moles is roughly the same, the empirical formula for the compound is <em>CO</em>.

4 0
3 years ago
Which class of organic compound is most likely to be used in anesthetics?
lubasha [3.4K]

The class of organic compound is most likely to be used in anesthetics is ethers. The first ether that is used in anesthetics is diethyl ether .

In general anesthetic works on brain, and produce unconsciousness and insensitivity to feel pain or anything. Generally it lowers the sensitivity of the organs.  Diethyl ether is the first ether that is used as anesthetics.


8 0
4 years ago
Read 2 more answers
How many molecules of nitrogen gas are in 15 L of nitrogen gas
uranmaximum [27]

Answer:

3.98\times 10^{23}

Explanation:

Since we have $N_A$ number of molecules in 22.4 L, in 15 L we have:

$\frac{15}{22.4}\times 6.023 \times 10^{23}$ molecules.

5 0
3 years ago
Relacione la teoria cuantica con el numero cuantico correspondite
Xelga [282]
“relate the quantum theory to the corresponding quantum number” what is your quantum number
7 0
3 years ago
Other questions:
  • _______ is the current condition in the troposphere. (2 points)
    14·2 answers
  • Why is DDT a dangerous environmental pollutant?
    12·2 answers
  • Which subctant will not conduct electricity a Aluminium b copper c plastic d steel​
    11·2 answers
  • Please Help!!!!!!!
    15·1 answer
  • Describe how replacing one light on a holiday string of bulbs with a "blinking" light would cause all the lights in the string t
    9·1 answer
  • Are these statements true or false? 1. Two sigma bonds comprise a double bond. 2. A triple bond consists of one pi bond and two
    14·1 answer
  • Which of the following is a state of matter ? particle,thermal,plasma, and diffuse ?
    12·2 answers
  • Transformation of energy through digestion that helps maintain body temperatures
    14·1 answer
  • Will give brainliest
    15·1 answer
  • I need help with this
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!