We are given the mass spectrum data for this compound which has a molecular ion peak of m+ = 91.043 m/z. When we have an m+ peak that is an odd number, that suggests that there are an odd number of nitrogens, in this case we'll assume 1 nitrogen atom to start. Nitrogen has a mass of 14 so we will substract that from our initial value.
91- 14 (1N) = 77 m/z
We are also told that there are carbon, hydrogens and oxygens present, so we will assume there is at least one oxygen which has a mass of 16 and subtract that value.
77 - 16 (1 O) = 61 m/z
Now we will try to get as close as possible to the remaining mass with carbons that has a mass of 12, and fill the remaining mass with hydrogens that have a mass of 1.
61 / 12 = 5
5 x 12 = 60
61 - 60 (5 C) = 1 m/z and this leaves us with 1 H.
The current formula would be C₅HON, but this structure is impossible since we do not have enough hydrogens to satisfy the carbons. So we can try to use 4 carbons instead and fill the rest with hydrogens.
4 x 12 = 48
61 - 48 (4 C) = 13 m/z and this leaves us with 13 H.
The current formula would be C₄H₁₃ON. The most hydrogens we can have in a compound is 2n+2 where n is the number of carbons. So with 4 carbons the most hydrogens we could have is 10. Therefore, our formula has too many hydrogens and also cannot work. So we cannot make up the remaining mass with carbons and hydrogens, therefore, we should add another oxygen before working with carbons and hydrogens.
61 - 16 (1 O) = 45 m/z
45/ 12 = 3.75
3 x 12 = 36
45 - 36 (3 C) = 9 m/z which gives us 9 hydrogens left.
The current formula is now C₃H₉O₂N. To test if this formula works we can calculate the double bond equivalents (DBE), also known as degrees of unsaturation.
DBE = C - H/2 + N/2 + 1 = 3 - (9/2) + (1/2) + 1 = 0
A value of 0 DBE tells us that there are no double bonds in this molecule but that the formula is a possibility so:
C₃H₉O₂N = 91 m/z
Dalton's Theory:
1. all elements is composed of very small particles called atoms
<span>2. Atoms of a given element are identical in size, mass, and other properties; atoms of different elements differ in size, mass, and properties. </span>
3. atoms cannot be divided, created, or destroyed
<span>4. Atoms of different elements combine in simple whole number ratios to form chemical compounds </span>
<span>5. In chemical reactions, atoms are combined, separated, or rearranged.
</span>So as we can see; the first two sentences are correct while the third sentence not supported in the theory
<span>3 ZnBr2 (aq) + 2 Al (s) ==> 2 AlBr3 (aq) + 3 Zn (s)
The unbalanced equation is:
ZnBr2 (aq) + Al (s) ==> AlBr3 (aq) +Zn (s)
First, count the atoms of each element on each side of the equation:
Zn 1,1
Br 2,3
Al 1,1
The zinc and aluminum, but the bromine doesn't match with 2 and 3. So look for the least common multiple of 2 and 3 which is 6 and adjust the quantities on both sides to have 6 bromine atoms on both sides. Do this by having 3 zinc bromide on the left and 2 aluminum bromide on the right, getting:
3 ZnBr2 (aq) + Al (s) ==> 2 AlBr3 (aq) +Zn (s)
Now check the atom counts again for both sides:
Zn 3,1
Br 6,6
Al 1,2
Now bromine matches, but zinc and aluminum doesn't. But it's easy enough to add an extra aluminum to the left and 2 more zinc to the right. Giving:
3 ZnBr2 (aq) + 2 Al (s) ==> 2 AlBr3 (aq) + 3 Zn (s)
Now check the atom counts again:
Zn 3,3
Br 6,6
Al 2,2
And they match. So the balanced equation is:
3 ZnBr2 (aq) + 2 Al (s) ==> 2 AlBr3 (aq) + 3 Zn (s)</span>
Initially, you should know that what the text describes is a test used to analyze and identify some metal ions in a compound. This is the principle behind it: first the ions are excited, by the heat of the flame, they absorb energy and the electrons are promoted to higher energy levels; after this, the electrons will fall back down to lower energy levels releasing energy as light. The flame color that is seen is related with the frequency of the light emitted.
Here is the text with the right words.
In this experiment, the metal cations in the solutions were initially in
the ground state. When placed in the flame, the metals then (absorbed)
energy as (heat). When this occurred,
electrons made transitions from (low) energy levels to (high)
energy levels. The metals were then in the (excited) state. The
electrons in these metals then made transitions from (high) energy
levels to (low) energy levels, resulting in the (emission) of energy as (EM radiation).
