Explanation:
A period 3 element is one of the chemical elements in the third row (or period) of the periodic table of the chemical elements. The periodic table is laid out in rows to illustrate recurring (periodic) trends in the chemical behaviour of the elements as their atomic number increases: a new row is begun when the periodic table skips a row and a chemical behaviour begins to repeat, meaning that elements with similar behaviour fall into the same vertical columns. The third period contains eight elements: sodium, magnesium, aluminium, silicon, phosphorus, sulfur, chlorine, and argon. The first two, sodium and magnesium, are members of the s-block of the periodic table, while the others are members of the p-block. All of the period 3 elements occur in nature and have at least one stable isotope.[1]
Answer:
It s a pure substance.
Explanation:
Can only be separated into its different elements by chemical means because it is connected by a chemical bond.
☃️ Chemical formulae ➝ 
<h3>
<u>How to find?</u></h3>
For solving this question, We need to know how to find moles of solution or any substance if a certain weight is given.
<h3>
<u>Solution:</u></h3>
Atomic weight of elements:
Ca = 40
C = 12
O = 16
❍ Molecular weight of
= 40 + 12 + 3 × 16
= 52 + 48
= 100 g/mol
❍ Given weight: 10 g
Then, no. of moles,
⇛ No. of moles = 10 g / 100 g mol‐¹
⇛ No. of moles = 0.1 moles
☄ No. of moles of Calcium carbonate in that substance = <u>0.1 moles</u>
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Can u show picture because without picture I don’t understand the problem u trying to explain
The complete question is;
Calculate the empirical formula for each of the following naturalflavors based on their elemental mass percent composition.
Q1)
methyl butyrate (component of apple taste andsmell): C -58.80 % H- 9.87 %
O -31.33.%Express your answer as a chemical formula.
Q2)
vanillin (responsible for the taste and smellof vanilla): C - 63.15% H- 5.30 %
O - 31.55%Express your answer as a chemical formula.
Q1)
empirical formula is the simplest ratio of whole number of elements in the compound. as the percentages have been given, lets calculate for 100 g of compound
C H O
mass 58.80 g 9.87 g 31.33
molar mass 12 g/mol 1 g/mol 16 g/mol
number of moles 58.80/12 9.87/1 31.33/16
= 4.9 =9.87 = 1.95
then divide number of moles by least number of moles - 1.95 in this case
4.9/1.95 = 2.51 9.87/1.95 = 5.06 1.95/1.95= 1
next multiply by 2 to get numbers that can be rounded off to whole numbers
2.51x2 = 5.02 5.06x2 = 10.12 1x2 = 2
when rounded off to the nearest whole number
C - 5
H - 10
O - 2
therefore empirical formula is C₅H₁₀O₂
Q2) for this too since elemental composition has been given in percentages lets calculate for 100 g of compound
C H O
mass 63.15 g 5.30 g 31.55 g
molar mass 12 g/mol 1 g/mol 16 g/mol
number of moles 63.15/12 5.30/1 31.55/16
=5.26 =5.30 =1.97
divide the number of moles by the least number of moles - 1.97
5.26/1.97 5.30/1.97 1.97/1.97
=2.67 = 2.69 = 1
multiply each by 3 to get numbers that can be rounded off to whole numbers
2.67x3 = 8.01 2.69x3 = 8.07 1x3 = 3
rounded off to the nearest whole numbers
C - 8
H - 8
O - 3
empirical formula = C₈H₈O₃
