The nitrate of metal $\mathbf{M}$ reacts with 1,4-benzenedicarboxylic acid (abbreviated as $\mathrm{H_{2}BDC}$) in a mixed system of N,N-dimethylformamide (abbreviated as $\mathrm{DMF}$) and anhydrous ethanol to yield a valuable material $\mathbf{A}$. Single-crystal X-ray diffraction analysis of $\mathbf{A}$ reveals that $\mathbf{A}$ belongs to the orthorhombic crystal system, with unit cell parameters $\mathrm{a=1923.01 \ \mathrm{pm}}$, $\mathrm{b=723.87 \ \mathrm{pm}}$, $\mathrm{c=884.82 \ \mathrm{pm}}$, $\mathrm{Z=4}$, and a density of $\mathrm{\rho =1.597 \ \mathrm{g / c m ^ { 3 } }}$.

The framework structure of $\mathbf{A}$ primarily consists of $\mathbf{M}$ and $\mathrm{BDC^{2-}}$, with solvent molecules incorporated in the structure. Elemental analysis indicates that $\mathbf{A}$ contains five elements: $\mathbf{M}$, $\mathrm{C}$, $\mathrm{H}$, $\mathrm{O}$, and $\mathrm{N}$, with an atomic ratio of $\mathbf{M}$ to $\mathrm{C}$ of 1:11. When $\mathbf{A}$ is subjected to high-temperature oxidation in air, nanoparticles of a common oxide of $\mathbf{M}$ are obtained, and the mass fraction of $\mathbf{M}$ in this common oxide is approximately 70%.

Based on the above information, determine $\mathbf{M}$.
金属  $\mathbf{M}$  的硝酸盐与1,4-对苯二甲酸 (简写为 $\mathrm{H_{2}BDC}$ ) 在N,N-二甲基甲酰 (简写为 $\mathrm{DMF}$ ) 和无水乙醇的混合体系中反应，得到一种有价值材料$\mathbf{A}$。对$\mathbf{A}$的单晶X射线衍射分析表明，$\mathbf{A}$属于正交晶系，晶胞参数 $\mathrm{a=1923.01 \ \mathrm{pm}}$，$\mathrm{b=723.87 \ \mathrm{pm}}$， $\mathrm{c=884.82 \ \mathrm{pm}}$， $\mathrm{Z=4}$，且密度 $\mathrm{\rho =1.597 \ \mathrm{g / c m ^ { 3 } }}$。

$\mathbf{A}$的骨架结构主体由$\mathbf{M}$ 和$\mathrm{BDC^{2-}}$组成，结构中带有溶剂分子。元素分析表明，$\mathbf{A}$ 中含有$\mathbf{M}$，$\mathrm{C}$， $\mathrm{H}$，$\mathrm{O}$和$\mathrm{N}$五种元素，且$\mathbf{M}$ 与$\mathrm{C}$的原子比为1:11。若将$\mathbf{A}$置于空气氛围下高温氧化，则可得到 $\mathbf{M}$ 一常见氧化物的纳米颗粒，且该常见氧化物中$\mathbf{M}$ 的质量分数约70 %。

根据以上信息，请判断$\mathbf{M}$。
Co
Ni
Cu
Ca
Fe
Na
Co
Ni
Cu
Ca
Fe
Na
Based on the crystallographic data of material $\mathbf{A}$, the chemical formula of $\mathbf{A}$ can be inferred. From $\mathrm{Mr=\frac{\rho N_AV}{Z}}$, substituting the data yields a molecular weight of $\mathrm{296.03\ \ g/mol}$ for this material.  

<Checkpoint>[$\mathrm{Mr(\mathbf{A})=296.03 \ \ g/mol}$](1)</Checkpoint>  

The problem states that the ratio of $\mathbf{M}$ to $\mathrm{C}$ in $\mathbf{A}$ is 1:11, which can be used to deduce the chemical composition of $\mathbf{A}$. Assuming for now that the chemical formula of $\mathbf{A}$ contains only one $\mathbf{M}$, this would correspond to 11 $\mathrm{C}$ atoms. One $\mathrm{(BDC)^{2-}}$ contains 8 C atoms, so the formula should include only one $\mathrm{(BDC)^{2-}}$. This leaves 3 C atoms unaccounted for, which matches the carbon count of one DMF molecule. If the chemical formula is $\mathrm{\mathbf{M}(BDC)(DMF)}$, calculating the molecular weight gives a metal molecular weight of $\mathrm{58.80\ \ g/mol}$.  

<Checkpoint>[$\mathrm{Mr(\mathbf{M})=58.80\ \ g/mol}$](1)</Checkpoint>  

According to the periodic table, the metals that meet this requirement could be Co or Ni. Based on the given conditions, when this metal is oxidized at high temperatures in air, it forms a common oxide with a mass fraction of the metal of approximately 70%. Possible oxides of Co include $\mathrm{CoO}$, $\mathrm{Co_2O_3}$, and $\mathrm{Co_3O_4}$, while the oxide of Ni is NiO. Considering the mass fraction, the metal is Co.  

<Checkpoint>[Based on the mass fraction, the metal is Co](1)</Checkpoint>
可以根据材料$\mathbf{A}$的晶体学数据，对于$\mathbf{A}$的化学式进行推断。由$\mathrm{Mr=\frac{\rho N_AV}{Z}}$，代入数据得该材料分子量为$\mathrm{296.03\ \ g/mol}$。

<Checkpoint>[$\mathrm{Mr(\mathbf{A})=296.03 \ \ g/mol}$](1)</Checkpoint>

题目中给出，$\mathbf{A}$中的$\mathbf{M}$和$\mathrm{C}$比例为1:11，可以据此对$\mathbf{A}$的化学组成进行推测。不妨先假定$\mathbf{A}$的化学式中仅含一个$\mathbf{M}$，此时对应11个$\mathrm{C}$。一个$\mathrm{(BDC)^{2-}}$包含8个C，故应仅含有一个$\mathrm{(BDC)^{2-}}$。此时还有3个C无去向，正好与一个DMF的碳数对应。若化学式为$\mathrm{\mathbf{M}(BDC)(DMF)}$，对分子量进行计算，该金属分子量为$\mathrm{58.
80\ \ g/mol}$。

<Checkpoint>[$\mathrm{Mr(\mathbf{M})=58.80\ \ g/mol}$](1)</Checkpoint>

根据元素周期表，满足该要求的金属可能为Co或Ni。根据题干，该金属在空气中高温氧化，会生成一种金属的质量分数约70 %的常见氧化物。Co的氧化物可能为$\mathrm{CoO}$、$\mathrm{Co_2O_3}$、$\mathrm{Co_3O_4}$，Ni的氧化物为NiO。考虑质量分数，该金属为Co。

<Checkpoint>[根据质量分数，该金属为Co](1)</Checkpoint>